Answers
Answer:
Explanation:
to begin, three steps are highlighted below which are used to write the Lewis structures of the given compounds, viz;
sum the variance of all the atomuse a pair of electrons to form a bond between each pair of bound atoms lastly arrange the remaining electrons so as to satisfy the rule for hydrogen and the octet rule for the second-row elements.the image below gives a step by step explanation as to answering this question.
i hope this was helpful, cheers.
Drawing Lewis structures entails representing the arrangement of electrons in a molecule, observing the octet rule, and assigning formal charges. The formal charge is the hypothetical charge an atom would possess if electrons in bonds are evenly distributed. The negative formal charges are preferably located on the most electronegative atoms in the molecule or ion when there are multiple possible structures.
Explanation:Drawing Lewis structures and assigning formal charges requires an understanding of the octet rule and the nature of the molecules. Lewis structures depict the arrangement of electrons in a molecule, particularly illustrating the bonding between atoms and the lone pairs of electrons that may exist. The octet rule suggests that atoms are stable when their outermost (valence) shell is full, typically with eight electrons.
The formal charge on an atom in a molecule is the hypothetical charge the atom would have if we could redistribute the electrons in the bonds evenly between the atoms. We calculate formal charge as follows: Formal Charge = [# of valence electrons on atom] – [non-bonded electrons + number of bonds]. Lewis structures are most reliable when adjacent formal charges are zero or of the opposite sign, and if there are several possible structures for a molecule or ion, the one with the negative formal charges on the more electronegative atoms is often the most accurate.
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Related Questions
A chemist makes of magnesium fluoride working solution by adding distilled water to of a stock solution of magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to significant digits.
Answers
The question is incomplete, here is the complete question:
A chemist makes 600. mL of magnesium fluoride working solution by adding distilled water to 230. mL of a stock solution of 0.00154 mol/L magnesium fluoride in water. Calculate the concentration of the chemist's working solution. Round your answer to 3 significant digits.
Answer: The concentration of chemist's working solution is [tex]5.90\times 10^{-4}M[/tex]
Explanation:
To calculate the molarity of the diluted solution (chemist's working solution), we use the equation:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of the stock magnesium fluoride solution
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of chemist's magnesium fluoride solution
We are given:
[tex]M_1=0.00154M\\V_1=230mL\\M_2=?M\\V_2=600mL[/tex]
Putting values in above equation, we get:
[tex]0.00154\times 230=M_2\times 600\\\\M_2=\frac{0.00154\times 230}{600}=5.90\times 10^{-4}M[/tex]
Hence, the concentration of chemist's working solution is [tex]5.90\times 10^{-4}M[/tex]
A mineral sample is obtained from a region of the country that has high arsenic contamination. An elemental analysis yields the following elemental composition:
Element Atomic Weight (g/mol) Percent Composition
Ca 40.078 22.3%
As 74.9216 41.6%
O 15.9994 35.6%
H 1.00794 60%
What is the empirical formula of this mineral?
Answers
Answer:
CaAsHO₄
Explanation:
The data has a mistake in one of the values there. I believe the mistake is on the hydrogen. So, I'm going to assume the value of Hydrogen is 0.6%, so the total percent composition would be 100.1% (Something better). All you have to do is replace the correct value of H (or the value with the mistaken option) and do the same procedure.
Now, to calculate the empirical formula, we can do this in three steps.
Step 1. Calculate the amount in moles of each element.
In these case, we just divide the percent composition with the molar mass of each one of them:
Ca: 22.3 / 40.078 = 0.5564
As: 41.6 / 74.9216 = 0.5552
O: 35.6 / 15.9994 = 2.2251
H: 0.6 / 1.00794 = 0.5953
Now that we have done this, let's calculate the ratio of mole of each of them. This is doing dividing the smallest number of mole between each of the moles there. In this case, the moles of As are the smallest so:
Ca: 0.5564/0.5552 = 1.0022
As: 0.5552/0.5552 = 1
O: 2.2251/0.5552 = 4.0077
H: 0.5953/0.5552 = 1.0722
Now, we round those numbers, and that will give us the number of atoms of each element in the empirical formula
Step 3. Write the empirical formula with the rounded numbers obtained
In this case we will have:
Ca: 1
As: 1
O: 4
H: 1
The empirical formula would have to be:
CaAsHO₄
2-19 Compare and contrast metallic and covalent primary bonds in terms of
(a) the nature of the bond,
(b) the valence of the atoms involved, and
(c) the ductility of the materials bonded in these ways.
Answers
While metallic bonds have strong attractions and ionic bonds involve the transfer and acceptance of electrons from the valence shell, covalent bonds involve sharing electrons.
What are chemical bonds?Chemical bonds are defined as a link made by two surfaces or items that have been brought together, often by heat, pressure, or an adhesive agent. Chemical bonds can have many different types, but covalent and ionic bonds are the most well-known. When one atom has less energy, the other has enough thanks to these bonds. Atoms are held together by the force of attraction, which enables the electrons to unite in a bond.
In contrast to covalent bonding, which involves atoms sharing their additional electrons locally, metallic bonding involves all atoms giving off their extra electrons and forming a sea of them. The ability of an atom to stick together and fill the orbits of its outermost electrons in order to arrange itself in a form that is the most stable.
Thus, while metallic bonds have strong attractions and ionic bonds involve the transfer and acceptance of electrons from the valence shell, covalent bonds involve sharing electrons.
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Barium has a body-centered cubic structure. If the atomic radius of barium is 222 pm, calculate the density of solid barium?
Answers
Answer: density =3.377g/cm³
Explanation:
Density =( molecular weight × effective number of atoms per unit cell) / (volume of unit cell × avogadro constant)
D= (M ×n) /(V×A)
M= 137g/mol
n= 2 (For BCC)
V=a³ , where a= 4r/√3
a= (4×222)/√3
a=512.69pm
a= 512.69×10^-10cm
V= ( 512.69×10^-10)^3
V= 1.3476×10^-22cm³
D= (137×2)/(1.3476×10^-22 × 6.02^23)
D= 3.377g/cm³
Therefore the density of barium is 3.377g/cm³
Given:
Radius of barium,
R = 222 pmWe know,
→ [tex]a = \frac{4r}{\sqrt{3} }[/tex]
[tex]= \frac{4\times 222}{\sqrt{3} }[/tex]
[tex]= 512.69 \ pm[/tex]
[tex]= 512.69\times 10^{-10} \ cm[/tex]
Now,
→ [tex]V = a^3[/tex]
[tex]= (512.69\times 10^{-10})^3[/tex]
[tex]= 1.3476\times 10^{-22} \ cm^3[/tex]
hence,
The density will be:
→ [tex]D = \frac{M\times n}{V\times A}[/tex]
[tex]= \frac{137\times 2}{1.3476\times 10^{-22}\times 6.02\times 10^{23}}[/tex]
[tex]= 3.377 \ g/cm^3[/tex]
Thus the response above is right.
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What are two important quantum-mechanical concepts associated with the Bohr model of the atom?
Answers
Explanation:
The important quantum-mechanical concepts associated with the Bohr model of atom are :
1. Electrons are nothing but particles that revolve around the nucleus in discrete orbitals.
2. Energy is associated with each orbital is quantised. Meaning electron in each shell will have energy in multiple of a fixed quanta.
Final answer:
The Bohr model introduced two key concepts of quantum mechanics: the quantization of energy levels and the quantization of angular momentum, both crucial for understanding atomic structure and the emission spectra of atoms.
Explanation:
The Bohr model of the atom represents an early understanding of quantum mechanics, specifically applicable to the hydrogen atom. It introduced two fundamental quantum-mechanical concepts that underpin the modern understanding of atomic structure and behavior, despite its ultimate replacement by more comprehensive theories.
Quantized Energy Levels
The first important concept is the quantization of energy levels of electrons within an atom. According to the Bohr model, electrons can only occupy certain allowed energy levels, and transitions between these levels result in the absorption or emission of photons with specific energies. This concept matched the observed discrete spectra of hydrogen, although the model assumed circular orbits for electron paths, which was later found to be incorrect.
Quantized Angular Momentum
The second concept is the quantization of angular momentum. Bohr proposed that the angular momentum of an electron in orbit is quantized and can only take on certain values, each associated with a different orbit. Although Bohr's assumptions about specific, circular orbits were eventually supplanted by the probabilistic nature of electron positioning in quantum mechanics, the idea that angular momentum is quantized remains a fundamental aspect of quantum theory.
Suppose you have a bucket of sand containing 2.5 billion grains of sand (2.5 x 10 grains). Most of the grains of sand are white, but there are some brown grains mixed evenly throughout. Number If the concentration of brown sand is 2.0%, how many grains of brown sand are in the bucket? brown grains Number If the concentration of brown sand is 2.0 ppm I how many grains of brown sand are in the bucket?
Answers
Answer:
a. 5.0 x 10⁷ brown grains = 50 million
b. 5.0 x 10³ brown grains = 5000
Explanation:
The concentration of 2 % brown sand means we have for every 100 grains of sand 2 are brown.
We need to calculate the number of brown sand in the bucket as follows:
= 2.5 x 10⁹ billion grains of sand x (2 brown grains/ 100 grains of sand)
= 5.0 x 10⁷ brown grains
Likewise if the concentration of brown sand is 2.0 ppm, it mean that we have 2 brown grain per every million grains of sand.
= 2.5 x 10⁹ billion grains of sand x ( 2.0 brown grains/10⁶ grains of sand )
= 5.0 x 10³ brown grains
The answers make sense since a concentration of 1 part per million is ten thousandths of a 1 percent
If the concentration of brown sand is 2.0%, there are 5 x 10^7 brown sand grains in the bucket.
If the concentration of brown sand is 2.0 ppm, there are also 5 x 10^7 brown sand grains in the bucket.The mass and volume of floating objects can vary widely depending on the specific properties of the objects and the fluid they are floating in (usually a liquid, such as water). However, there are some general principles that hold true for all floating objects:**Mass Displacement:** When an object floats in a fluid, it displaces an amount of the fluid equal to its own weight. This principle is described by Archimedes' principle. Therefore, the mass of the fluid displaced by the object is equal to the mass of the object itself. In other words, the object's mass is balanced by the mass of the fluid it displaces.**Volume and Buoyancy:** The volume of a floating object is typically less than or equal to the volume of the fluid it displaces. If the object's volume were greater than the displaced fluid's volume, it would sink. Therefore, floating objects have a volume that allows them to displace a sufficient amount of fluid to support their weight.The mass of a floating object is typically equal to the mass of the fluid it displaces, while the object's volume is less than or equal to the volume of the displaced fluid. These principles ensure that the object remains buoyant and does not sink in the fluid.For more such questions on concentration
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.What is the composition in weight percent of an alloy that contains 98 g tin and 65 g lead?What is the composition in atom percent?
Answers
Answer:
See explanation below for answer
Explanation:
We'll do this question in two parts:
a) Weight percent of the alloy:
This is pretty easy, all we have to do is to sum the mass of each element in the alloy and then, divide each mass between the total to get the percentage in weight. In other words:
%W = m/mtotal * 100
The total mass:
mtotal = 98 + 65 = 163 g
%Sn = 98/163 * 100 = 60.12%
%Pb = 65/163 * 100 = 39.88%
b) atom percent of the alloy:
In this case, we do something similar that in part a) with the difference that instead of using mass, we will use moles of each element. The molar mass of Tin and Lead reported are 118.71 g/mol and 207.2 g/mol. The expression to use will be:
C = n/ntotal * 100
so for each element, the moles are:
moles Sn = 98/118.71 = 0.8255 moles
moles Pb = 65/207.2 = 0.3137 moles
the total moles:
ntotal = 0.8255 + 0.3137 = 1.1392 moles
So the composition in atoms for each element is:
C Sn = 0.8255/1.1392 * 100 = 72.46%
C Pb = 0.3137/1.1392 * 100 = 27.54%
The weight percent of tin in an alloy of 98 g tin and 65 g lead is 60.12%, and the weight percent of lead is 39.88%. The atom percent of tin is 71.26% and the atom percent of lead is 28.74%.
Explanation:The question is asking for the composition in weight percent of an alloy that contains 98 g tin and 65 g lead, and also its composition in atom percent.
To find the weight percent of a component in an alloy, the formula is: (mass of component/total mass) x 100%. Therefore, the weight percent of tin in this alloy is: (98/(98+65)) x 100% = 60.12% and the weight percent of lead is 100% - 60.12% = 39.88%.
The composition in atom percent refers to the number of atoms of a component over the total number of atoms. Lead and tin both have different atomic masses, so we can't use the same weights. The atomic mass of tin is 118.71 g/mol and lead is 207.2 g/mol. Therefore, the atom percent of tin is: (98/118.71)/(98/118.71 + 65/207.2) x 100% = 71.26%. The atom percent of lead is 100% - 71.26% = 28.74%.
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Calculate the amount of heat needed to boil 183.g of ethanol ( CH3CH2OH ), beginning from a temperature of 33.9°C . Round your answer to 3 significant digits. Also, be sure your answer contains a unit symbol.
Answers
To calculate the amount of heat needed to boil ethanol, use the formula Q = m * c * ΔT. Convert the temperature to Kelvin, calculate the change in temperature, and then apply the formula to find the amount of heat. The amount of heat needed to boil 183.g of ethanol is 20 kJ.
Explanation:To calculate the amount of heat needed to boil ethanol, we can use the equation:
Q = m * c * ΔT
Where Q is the amount of heat, m is the mass of ethanol, c is the specific heat capacity of ethanol, and ΔT is the change in temperature.
First, we need to calculate the change in temperature:
Convert the temperature from Celsius to Kelvin: 33.9°C + 273.15 = 306.05 KNext, subtract the boiling point of ethanol (78.4°C = 351.55 K) from the initial temperature: ΔT = 351.55 K - 306.05 K = 45.50 KNow we can calculate the amount of heat:
Find the specific heat capacity of ethanol: 2.44 J/g°CApply the formula: Q = 183.g * 2.44 J/g°C * 45.50 KFinally, round the answer to 3 significant digits: Q = 20000 J = 20 kJTherefore, the amount of heat needed to boil 183.g of ethanol is 20 kJ.
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The equilibrium A<=' B + Cat 25°C is subjected to a temperature jump which slightly increases the concentrations ofB and C. The measured relaxation time is 3.0 µs. The equilibrium constant for the system is 2.0 x 10 '" 16 at the newtemperature. and th e equilibrium concentrations ofBand C then are both 0.20 mmol dm-). Calculate the rate constants fo r the forward and reverse stepsgiven that the forward s tep is first-order in A, and the reverse step is first-order in both Band C.
Answers
The rate constants for the forward and reverse steps given that the forward step is first-order in A, and the reverse step is first-order in both B and C is 6.67 x 10⁻⁹ sec⁻¹.
What is rate constant?Rate constant is defined as the proportionality constant in the equation expressing the association between a chemical reaction's pace and its constituents' concentrations. The dependence of the molar concentration of the reactants on the rate of reaction can be determined by knowing the rate constant.
Given rt = 3μs = 3 x 10⁻⁶ s
K eq = 2 x 10⁻¹⁶
B = 2 x 10⁻⁴
C = 2 x 10⁻⁴
rt = 1 / Kf + Kr
3 x 10⁻⁶ = 1 / Kf + Kr ------- ( 1 )
K eq = Kf / Kr
2 x 10⁻¹⁶ = Kf / Kr ----------- ( 2 )
From equation 1 and 2
3 x 10⁻⁶ = 1 / Kf + Kf / 2 x 10⁻¹⁶
3 x 10⁻⁶ = 1 / 2 x 10⁻¹⁶ x Kf + Kf / 2 x 10⁻¹⁶
Kf = 2 x 10⁻¹⁶ / 3 x 10⁻⁶ ( 2 x 10⁻¹⁶ + 1)
Kf = 6.67 x 10⁻⁹ sec⁻¹
Thus, the rate constants for the forward and reverse steps given that the forward step is first-order in A, and the reverse step is first-order in both B and C is 6.67 x 10⁻⁹ sec⁻¹.
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Which of the following statements is true? I. Carbon forms covalent rather than ionic bonds. II. Carbon has an electronegativity value that is midway between that of the most metallic element and the most nonmetallic element. III. Carbon forms four bonds in virtually all its compounds. IV. A Si-Si bond is much stronger than a C-C bond. V. Relatively little heat is released when a C chain reacts and one bond replaces the other. Group of answer choices I, II, II, V I only II only III only IV only V only I and II I and III I and IV I and V II and III II and IV II and V III and IV III and V
Answers
Explanation:
Atomic number of carbon is 6 and its electronic distribution is 2, 4. As there are 4 valence electrons present in a carbon atom. Hence, it belongs to group 14 of the periodic table.
So, being a non-metal carbon atom tends to form covalent bonds. Also, the electronegativity value of carbon atom is mid-way between those of metals and non-metals.
In order to complete its octet, carbon atom tends to gain or share its 4 valence electrons. Therefore, it forms four bonds in all its compounds.
The Si-Si bond will be less stronger than C-C bond due to the larger size of silicon atom. As a result, there will be less overlapping between the silicon atoms due to which the bond formed will be weak in nature.
Whereas C-C bond is more stronger as a carbon atom is smaller in size as compared to silicon atom. As a result, more will be the overlapping between the carbon atoms. Hence, more stronger will be C-C bond than Si-Si bond.
As there will be covalent bonds present in a carbon chain and these bonds are weak as compared to ionic bonds. Hence, relatively little heat is released when a C chain reacts and one bond replaces the other.
Thus, we can conclude that out of the given options statements of I, II, III and V are true.
Calculate the radius of a platinum atom in cm, given that Pt has an FCC crystal structure, a density of 21.4 g/cm3, and an atomic weight of 195.08 g/mol.
Answers
Answer:
Radius of platinum atom is [tex]1,38\times10^{-8} \ cm.[/tex]
Explanation:
We know, Density in solid state is also given by , [tex]\rho=\dfrac{Z\times M}{N_A\times a^3}[/tex]
Here, Z= total no of atom or molecule per unit cell. ( Z = 4 for FCC structure )
M = atomic weight = 195.08 gm/mol.
N_A= Avogadro's number = [tex]6.023\times 10^{23} .[/tex]
a= size of side of cube.
Also for, FCC crystal [tex]a=\sqrt2\times 2r[/tex] ....1
Since,[tex]\rho=\dfrac{Z\times M}{N_A\times a^3}\\[/tex]
Therefore, [tex]a^3=\dfrac{Z\times M}{N_A\times \rho}[/tex]
Putting all values , [tex]a=3.9\times 10^{-8} \ cm.[/tex]
Putting value of a in equation 1 we get,
[tex]r=\dfrac{a}{\sqrt2\times2}=1.38\times 10^{-8} \ cm.[/tex]
Hence, this is the required solution.
The radius of a platinum atom can be found using the density, atomic weight, and the nature of the crystal structure. The calculation involves finding the volume of an atom from the atomic weight, density and Avogadro's number and converting this volume to a radius by considering the atom as a sphere.
Explanation:The radius of a platinum atom can be calculated using the density, atomic weight, and the information about the crystal structure (in this case, face-centered cubic or FCC). Platinum (Pt) has a density (d) of 21.4 g/cm3 and an atomic weight (AW) of 195.08 g/mol.
First, let's calculate the number of atoms (n) in the unit cell of an FCC structure. An FCC unit cell has 4 atoms. Avogadro's number (Na) is 6.022 x 1023 mol-1.
The volume (V) of an atom can be calculated as follows: V = AW / (d * n * Na).Now we will calculate the radius (r) from the volume, considering the atom as a sphere: r = ((3*V) / (4*π))1/3.Keep in mind that the values have to be converted to cm in the end.
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A 27.40 g sample of liquid mercury is initially at 158.30°C. If the sample is heated at constant pressure ( = 1 atm), kJ of energy are needed to raise the temperature of the sample to 376.20°C. How much energy in kJ is required?
Answers
Answer:
energy required is 0.247kJ
Explanation:
The formula to use is Energy = nRdT;
Where n is number of mole
R is the molar gas constant
dT is the change in temperature
n = reacting mass of mercury / molar mass of mercury = 27.4/200.59 = 0.137
dT = final temperature - initial temperature = 376.20 - 158.30 = 217.90K
R = 8.314Jper mol per Kelvin
Energy = 0.137 x 8.314 x 217.90 = 247.12J
Energy in kJ= 247.12/1000= 0.247kJ
CaCl_2 is often used to melt ice on sidewalks.
a) Could CaCl_2 melt ice at -33°C? Yes or no? Assume that the solubility of CaCl_2 at this temperature is 70.1 g CaCI_2/100.0 g of H2O and that the van't Hoff factor for a saturated solution of CaCl_2 is i = 2.5.
b) What is the minimum amount of CaCl_2 that would be required to melt ice on sidewalks at the above temperature? Use 100.0 g of water as the amount of solvent.
Answers
Answer:
a) No, it couldn't
b) 78.81 g
Explanation:
When a nonvolatile solute is added to a pure solvent, the melting point of the solvent must decrease, which is called cryoscopy. The temperature variation (melting point of pure solvent - the melting point of solution) can be calculated by:
ΔT = Kc.W.i
Where Kc is the cryoscopy constant (for water it's 1.86 °C/mol.kg), W is the molality, and i is the van't Hoff factor of the solute.
W =m1/m2*M1
Where m1 is the mass of the solute (in g), m2 is the mass of the solvent (in kg), and M1 is the molar mass of the solute (CaCl2 = 111.0 g/mol). The van't Hoff factor indicates how much of the solute ionizes in the solution.
a) The melting point of water is 0°C, so let's calculate the new melting point with the given information:
m1 = 70.1 g
m2 = 100 g = 0.1 kg
W = 70.1/(0.1*111) = 6.31 mol/kg
ΔT = 1.86*6.31*2.5
ΔT = 29.34°C
0 - T = 29.34
T = -29.34°C
Thus, at -33°C, the ice will not melt yet.
b) Let's then found out the value of m1 in this case:
ΔT = Kc.W.i
0 - (-33) = 1.86*W*2.5
4.65W = 33
W = 7.1 mol/kg
W = m1/M*m2
7.1 = m1/(111*0.1)
m1 = 78.81 g
Calcium chloride cannot melt the ice at -33°C. Cryoscopy is the method of determining a decrease in melting point due to dissolved substances.
What is cryoscopy?The decrease in the melting point of a pure solvent when a non-volatile solute is added to the solution.
[tex]\Delta T = K_c. W.i[/tex]
Where
[tex]K_c[/tex] - cryoscopy constant= 1.86 °C/mol.kg for water
[tex]W[/tex] - molality = 6.31 mol/kg
[tex]i[/tex] - Van't Hoff factor of the solute = 2.5
Put the values in the formula,
[tex]\Delta T = 1.86\times 6.31\times 2.5\\\\\Delta T = \rm \ 29.34^oC[/tex]
Therefore, Calcium chloride cannot melt the ice at -33°C.
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In a 4.00 L pressure cooker, water is brought to a boil. If the final temperature is 115 °C at 3.10 atm, how many moles of steam are in the cooker?
Answers
Answer: 0.39mol of the steam are in the cooker.
Explanation:Please see attachment for explanation
Given:
Volume, V = 4 LPressure, P = 3.10 atmTemperature, T = 115°C= 115+273
= 388 K
We know the relation,
→ [tex]PV = nRT[/tex]
or,
→ [tex]n = \frac{PV}{RT}[/tex]
By substituting the values, we get
[tex]= \frac{4\times 3.1}{0.082\times 388}[/tex]
[tex]= 0.39 \ mol[/tex]
[tex]= 0.39 \ mol[/tex]
Thus the above answer is correct.
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In the following list, only __________ is not an example of a chemical reaction.
A) burning a plastic water bottle
B) the production of hydrogen gas from water
C) the tarnishing of a copper penny
D) chopping a log into sawdust
E) charging a cellular phone
Answers
Answer: Thus chopping a log into sawdust is not an example of a chemical reaction.
Explanation:
Physical reaction is a reaction in which there is no rearrangement of atoms and thus no new substance is formed. There is only change in physical state of the substance.
Example: chopping a log into sawdust as the shape and size of wood changes.
Chemical reaction is a reaction in which there is rearrangement of atoms and thus new substance is formed. There may or may not be a change in physical state.
Example: burning a plastic water bottle : as the plastic would combine with oxygen to give oxidised products.
the production of hydrogen gas from water: the chemical composition changes
the tarnishing of a copper penny: oxidation of copper takes place to give copper oxide.
charging a cellular phone: the electrical energy is used for a chemical reaction
Chopping a log into sawdust is an example of a physical change because it changes the size not the composition of the substance.
What are chemical reactions?Chemical reactions involve such kind change that leads to the formation of a new product (compound).
For example- the production of hydrogen gas from water, the burning of plastic produce gases.
Physical reactions:These reactions refer to the rearrangements of the atoms in a substance but do not to the formation of a new product such as temperature, phase, size, etc.
For example- chopping a log into sawdust
Therefore, chopping a log into sawdust is an example of a physical change because it changes the size not the composition of the substance.
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Calculate the density, in g / mL , of a liquid that has a mass of 0.125 g and a volume of 0.000285 L
Answers
Answer:
ρ = 0.439 g/mL
Explanation:
Given data
Mass of the liquid: 0.125 gramsVolume of the liquid: 0.285 milliliters0.000285 L × (1000 mL/ 1 L) = 0.285 mL
The density (ρ) of a liquid is equal to its mass (m) divided by its volume (V).
ρ = m/V
ρ = 0.125 g/0.285 mL
ρ = 0.439 g/mL
The density of the liquid is 0.439 g/mL.
The density of a liquid, defined as mass per unit volume, with given mass of 0.125 grams and volume of 0.000285 liters, can be calculated using the formula: Density = mass/volume. After converting liters to milliliters, the density is approximately 0.439 g/mL.
Explanation:The question is asking for the calculation of the density of a liquid given its mass and volume. Density is a property of matter defined as mass per unit volume. It can be calculated using the formula: Density = mass/volume. In this case, the mass of the liquid is 0.125 grams and the volume is 0.000285 liters. But please note that the density should be presented in g/mL. Therefore, we need to convert the volume from liters to milliliters. 1 liter equals 1000 milliliters, so 0.000285 liters equals 0.285 milliliters. Using the formula, Density = 0.125 g / 0.285 mL which gives us a density of approximately 0.439 g/mL.
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Consider a general reaction
a) The ΔG° of the reaction is -4.140 kJ mol. Calculate the equilibrium constant for the reaction. (Assume a temperature of 25°C.)
b) What is ΔG at body temperature (37.0°C) if the concentration of A is 1.6 M and the concentration of B is 0.45 M?
Answers
Answer:
a) K = 5.3175
b) ΔG = 3.2694
Explanation:
a) ΔG° = - RT Ln K
∴ T = 25°C ≅ 298 K
∴ R = 8.314 E-3 KJ/K.mol
∴ ΔG° = - 4.140 KJ/mol
⇒ Ln K = - ( ΔG° ) / RT
⇒ Ln K = - ( -4.140 KJ/mol ) / (( 8.314 E-3 KJ/K.mol )( 298 K ))
⇒ Ln K = 1.671
⇒ K = 5.3175
b) A → B
∴ T = 37°C = 310 K
∴ [A] = 1.6 M
∴ [B] = 0.45 M
∴ K = [B] / [A]
⇒ K = (0.45 M)/(1.6 M)
⇒ K = 0.28125
⇒ Ln K = - 1.2685
∴ ΔG = - RT Ln K
⇒ ΔG = - ( 8.314 E-3 KJ/K.mol )( 310 K )( - 1.2685 )
⇒ ΔG = 3.2694
Ammonium carbamate (NH2COONH4) is a salt of carbamic acid that is found in the blood and urine of mammals. At 202°C, Kc = 0.008 for the following equilibrium: If 6.82 g of NH2COONH4 is put into a 0.5-L evacuated container, what is the total pressure (in atm) at equilibrium? Enter a number to 2 decimal places.
Answers
Answer:
P(total) = 46.08 atm
Explanation:
Given data:
Mass of NH₂COONH₄ = 6.8 g
Kc = 0.008
Temperature = 202°C = 202+273 = 475 K
Total pressure at equilibrium = ?
Solution:
Equilibrium equation:
NH₂COONH₄ ⇄ 2NH₃ + CO₂
Formula:
Kp = Kc(RT)³
Kp = 0.008 (0.0821 atm.L/mol.K )³(475K)³
Kp = 471.56
NH₃ CO₂
Initial concentration 0 0
Change in concentration 2x x
equilibrium concentration 2x x
AS
471.56 = 2x (x)
471.56 = 2x²
x² = 471.56 /2
x² = 235.78
x = 15.36
Pressure of ammonia = 2x = 2(15.36) = 30.72 atm
Pressure of carbon dioxide = x = 15.36 atm
Total pressure:
P(total) = P(NH₃) + P(CO₂)
P(total) = 30.72 atm + 15.36 atm
P(total) = 46.08 atm
write out the acid dissociation reaction for hydrochloric acid. b) calculate the pH of a solution of 5.0 x 10^-4 M HCl. c) write out the acid dissociation reaction for sodium hydroxide. d) calculate the pH of a solution of 7.0 x 10^-5 M NaOH.
Answers
Answer:
The answer to your question is below
Explanation:
a) Acid dissociation reaction for HCl
HCl(g) + H₂O(l) ⇒ H⁺¹ (aq) + Cl⁻¹ (aq)
b) pH = -log [H⁺¹]
Concentration = 5 x 10⁻⁴
pH = -log [5 x 10⁻⁴]
pH = 3.3
c) Acid dissociation reaction
NaOH(s) + H₂O ⇒ Na⁺¹(aq) + OH⁻¹(aq)
d) pH
pOH = -log [7 x 10⁻⁵]
pOH = 4.2
pH = 14 - 4.2
pH = 9.8
The acid dissociation and pH for HCl and NaOH were detailed. For HCl, the reaction and solution pH were HCl -> H+ + Cl- and 3.3 respectively. For NaOH, the dissociation and solution pH were NaOH -> Na+ + OH- and 9.8 respectively.
Explanation:a) The acid dissociation reaction for hydrochloric acid (HCl) is as follows:
HCl → H+ + Cl-
b) The pH of a solution of hydrochloric acid can be calculated using the formula:
pH = -log[H+].
As HCl is a strong acid, it will completely ionize in water, meaning the concentration of H+ ions equals the concentration HCl. Therefore, the pH for a 5.0 x 10^-4 M HCl solution is -log(5.0 x 10^-4) = 3.3.
c) Sodium hydroxide is a base, not an acid, so its ionization does not involve releasing protons into the solution, but rather attracting them. It has the following dissociation reaction:
NaOH → Na+ + OH-.
d) As for the pH of a solution of 7.0 x 10^-5 M NaOH, we first need to find the pOH using the formula pOH = -log[OH-], since NaOH being a strong base completely ionizes. This equals -log(7.0 x 10^-5) = 4.2. Given that pH and pOH are related by the expression pH + pOH = 14 at 25°C, we can solve for the pH of the NaOH solution = 14 - pOH = 14 - 4.2 = 9.8.
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A standard solution of 86% (w/w) H3PO4 (specific density = 1.71 g/mL) is to be diluted to get a1200 mL H3PO4 solution at 4.00 M. How many mL of the standard H3PO4 solution are required?
Answers
Answer: 319 ml
Explanation:
Given : 86 g of [tex]H_3PO_4[/tex] is dissolved in 100 g of solution.
Density of solution = 1.71 g/ml
Volume of solution=[tex]\frac{\text {mass of solution}}{\text {density of solution}}=\frac{100g}{1.71g/ml}=58.5ml[/tex]
moles of [tex]H_3PO_4=\frac{\text {given mass}}{\text {Molar mass}}=\frac{86g}{98g/mol}=0.88mol[/tex]
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution in L}}[/tex] .....(1)
Molarity of standard [tex]H_3PO_4[/tex] solution =[tex] \frac{0.88\times 1000}{58.5ml}=15.04M[/tex]
To calculate the volume of acid, we use the equation given by neutralization reaction:
[tex]M_1V_1=M_2V_2[/tex]
where,
[tex]M_1\text{ and }V_1[/tex] are the molarity and volume of standard acid which is [tex]H_3PO_4[/tex]
[tex]M_2\text{ and }V_2[/tex] are the molarity and volume of diluted acid which is [tex]H_3PO_4[/tex]
We are given:
[tex]M_1=15.04M\\V_1=?mL\\\\M_2=4.00M\\V_2=1200mL[/tex]
Putting values in above equation, we get:
[tex]15.04\times V_1=4.00\times 1200\\\\V_1=319mL[/tex]
Thus 319 ml of the standard [tex]H_3PO_4[/tex] solution are required.
To prepare a 1200 mL of 4.00 M H3PO4 solution from an 86% w/w H3PO4 solution with a specific density of 1.71 g/mL, 320 mL of the standard solution is required.
To determine how many milliliters of a standard 86% H3PO4 solution (with a specific density of 1.71 g/mL) are required to prepare a 1200 mL 4.00 M H3PO4 solution, we can use the concept of molarity and the equation M1V1 = M2V2, where M1 and V1 are the molarity and volume of the concentrated solution, and M2 and V2 are the molarity and volume of the diluted solution, respectively.
First, we need to calculate the molarity of the stock solution (86% H3PO4). We know that 86% (w/w) means there are 86 grams of H3PO4 in 100 grams of solution. As the density of the solution is 1.71 g/mL, we can find the volume that 100 grams of solution will occupy:
Volume = mass / density = 100 g / 1.71 g/mL = 58.48 mL
Now, we calculate the mols of H3PO4 in 86 grams:
Moles H3PO4 = mass (g) / molar mass (g/mol) = 86 g / 97.99 g/mol ≈ 0.877 mol
The molarity (M1) of the stock solution is:
Molarity = moles of solute / volume of solution (L) ≈ 0.877 mol / 0.05848 L ≈ 15.00 M
Next, using the dilution equation M1V1 = M2V2, where M2 = 4.00 M and V2 = 1200 mL (1.2 L), we can solve for V1:
V1 = (M2V2) / M1 = (4.00 M * 1.2 L) / 15.00 M = 0.32 L
Since we need the volume in milliliters:
V1 = 0.32 L * 1000 mL/L = 320 mL
Therefore, 320 mL of the standard 86% H3PO4 solution is required to prepare a 1200 mL of 4.00 M H3PO4 solution.
At the beginning of the industrial revolution in 1850, the CO2concentration was 280 ppm. Today, it is 410 ppm.
1. How much extra radiative forcing is the Earth’s surface receiving today, relative to 1850?
2. What is the equivalent temperature change? ΔF (Wm^(-2)) = α ln(C/C0), ΔT(K) = λ*ΔF, α = 5.35, λ = 0.8 per (Wm^(-2)).
Answers
Answer:
1. 2.04 W/m²
2. 1.63°C
Explanation:
The radiative force that the Earth receives comes from the Sun. When the Sun rays come to the surface, some of them are absorbed and then it is reflected in the space. The greenhouse gases (like CO2) blocks some of these rays, and then the surface stays warm. The excessive amount of these gases makes the surface warmer, which unbalance the climate on Earth.
1. The variation of the radiative forcing can be calculated based on the concentration of the CO2 by the equation:
ΔF = 5.35*ln(C/C0)
Where C is the final concentration, and C0 is the initial concentration.
ΔF = 5.35*ln(410/280)
ΔF = 2.04 W/m²
2. The temperature change in the Earth's surface caused by the variation of the radiative forcing can be calculated by:
ΔT = 0.8*ΔF
ΔT = 0.8*2.04
ΔT = 1.63 K = 1.63°C
The increase in CO2 concentration from 280 ppm to 410 ppm today results in an extra radiative forcing of 2.2 W/m² on Earth's surface. The equivalent temperature change is approximately 1.7°C.
Explanation:The increase in CO2 concentration from 280 ppm to 410 ppm today has resulted in an extra radiative forcing on Earth's surface. The radiative forcing is calculated using the formula ΔF (Wm^(-2)) = α ln(C/C0), where α = 5.35. By plugging in the values, the radiative forcing is approximately 2.2 W/m².
The equivalent temperature change can be calculated using the formula ΔT(K) = λ*ΔF, where λ = 0.8 per (Wm^(-2)). By multiplying the radiative forcing by the climate sensitivity, we get a temperature increase of about 1.7°C.
It is important to note that the temperature increase may catch up to a new equilibrium once oceans warm and ice melts, resulting in a higher temperature increase in the future.
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Typically, neutron stars are about 20 km in diameter and have around the same mass as our sun. What is a typical neutron star density in g/cm3?
Answers
Answer:
4.7 × 10¹⁴ g/cm³
Explanation:
The mass of the sun (and of the neutron star) is 1.98 × 10³⁰ kg = 1.98 × 10³³ g.
The diameter of the neutron star is 20 km and its radius is 1/2 × 20 km = 10 km = 10⁴ m = 10⁶ cm
If we consider the neutron star to be approximately spherical, we can calculate its volume using the following expression.
V = 4/3 × π × r³
V = 4/3 × π × (10⁶ cm)³
V = 4.19 × 10¹⁸ cm³
The density (ρ) is equal to the quotient between the mass and the volume.
ρ = 1.98 × 10³³ g / 4.19 × 10¹⁸ cm³ = 4.73 × 10¹⁴ g/cm³ ≈ 4.7 × 10¹⁴ g/cm³
Construct a simulated 1H NMR spectrum for ethyl acetate by dragging and dropping the appropriate splitting patterns into the boxes on the chemical shift baseline, and by dragging integration values into the small box above each signal. Items may be used more than once. Peak heights do not represent integration.
Answers
Answer:
Triplet at 1.15 ppm
Singlet at 2.02 ppm
Quartet at 4.10 ppm
Explanation:
In ethyl acetate there are three different sets of protons having different surrounding as compared to each other. The predicted ¹H-NMR of ethyl acetate is attached below and following are the signals shown by this compound,
(i) A singlet around 2.02 ppm is a characteristic peak shown by protons at carbon a. This is because there is no proton available on the adjacent carbon so there is no coupling.
(ii) A triplet around 1.15 ppm is a peak shown by protons at carbon c. This is because there are two proton present on the adjacent carbon so, according to (n+1) rule it will give a triplet peak.
(iii) A quartet around 4.10 ppm is a peak shown by protons at carbon b. This is because there are three proton present on the adjacent carbon so, according to (n+1) rule it will give a quartet peak. Also, as it is directly attached to oxygen atom hence, the chemical shift is downfielded or deshielded.
what is the kinetic energy of a particle of mass m moving through space with velocity v?
Answers
Answer:
Kinetic Energy of the particle is half the mass of the particle multiplied by the square of the velocity of the particle
Explanation:
Kinetic Energy is given by 1/2mv^2
Where, m is the mass of the object and v is the velocity of the object
Answer: kinetic energy is equal to 1/2mv^2
An aqueous solution is 18.0% by mass ammonia, NH3. The mole fraction of ammonia in the solution is:________
Answers
Answer:
0.189
Explanation:
Aqueous solution only has ammonia and water in it. Assume total mass of 100 g
Meaning mass of ammonia is 18% of 100 g = 18 g
Mass of water is 82% of 100 g = 82g
Number of moles of ammonia = mass / molar mass ammonia =18/17 = 1.059
Number of moles of water = mass / molar mass water = 82/18 =4.556
Mole fraction ammonia = 1.059 / (1.059+4.556) = 0.189
Final answer:
The mole fraction of ammonia in an 18.0% by mass aqueous solution of ammonia is calculated based on the ratio of moles of ammonia to total moles in the solution, resulting in a mole fraction of 0.188.
Explanation:
Mole fraction is the ratio of the moles of solute to the total moles in the solution. Given the percentage by mass, we can calculate the mole fraction of ammonia, assuming the mass of the solution is 100 grams for simplicity.
An 18.0% by mass solution means there are 18.0 grams of NH₃ and 82.0 grams of water (H₂O). To find the moles of each component, we divide their mass by their respective molar masses: NH₃ (17.031 g/mol) and H₂O (18.015 g/mol).
Moles of NH₃ = 18.0 g / 17.031 g/mol = 1.057 moles
Moles of H₂O = 82.0 g / 18.015 g/mol = 4.551 moles
The total moles in solution = 1.057 moles (NH₃) + 4.551 moles (H₂O) = 5.608 moles
The mole fraction of NH₃ = Moles of NH₃ / Total moles in solution = 1.057 / 5.608 = 0.188
Therefore, the mole fraction of ammonia in the solution is 0.188.
The octane rating of gasoline is a measure of the resistance to premature ignition, also known as engine knock. It is based on comparison to a mixture of 2,2,4-trimethylpentane, commonly known as isooctane (rating = 100), and heptane (rating = 0). An octane rating of 90 would compare to a mixture that is 90% isooctane and 10% heptane. Isooctane is a highly branched alkane with the condensed structure (CH3)3CCH2CH(CH3)2. Draw the complete struckture and skeletal struckture.
Answers
Answer: From the attached document
1. Complete structure
2. Skeletal structure
The complete structural formula for isooctane (2,2,4-trimethylpentane) is (CH3)3CCH2CH(CH3)2, while the skeletal formula is CH3-C-(CH3)2-CH2-CH(CH3)2. These represent its molecular structure in detail and a simplified version, respectively.
The complete structural formula and skeletal formula for 2,2,4-trimethylpentane, commonly known as isooctane, are as follows:
Complete Structural Formula:
In the complete structural formula, we represent each atom in the molecule explicitly. Isooctane has the following structure:
```
H
|
H - C - (CH3)2
| |
H - C - CH3
|
CH3
```
In this representation, each line represents a chemical bond, and we've labeled carbon (C) and hydrogen (H) atoms to show their positions in the molecule. Isooctane is highly branched, containing multiple methyl (CH3) groups attached to a central carbon atom.
Skeletal Formula:
The skeletal formula is a simplified way of representing the molecular structure. It focuses on the carbon framework, and hydrogen atoms bonded to carbon are usually omitted. Isooctane's skeletal formula is as follows:
```
CH3
|
CH3-C-(CH3)2
|
CH3
```
In this skeletal formula, each vertex represents a carbon atom, and the lines between them represent carbon-carbon bonds. The methyl (CH3) groups are indicated as branches off the carbon atoms.
Isooctane is an important component in the determination of gasoline's octane rating because it has a high resistance to engine knocking, making it a reference standard with an octane rating of 100. Gasoline blends with a higher percentage of isooctane are less prone to premature ignition and have higher octane ratings, indicating their suitability for high-performance engines.
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Calculate the mass of carbon dioxide produced from complete combustion of 16.6 g of octane, C8H18, in the following reaction. C8H18 O2 --> CO2 H2O (unbalanced)
A. 51.2 g CO2
B. 102 g CO2
C. 57.6 g CO2
D. 0.799 g CO2
E. 1.163 g CO2
Answers
Answer:
51.2g of CO2
Explanation:
The first step is to balance the reaction equation as shown in the solution attached. Without balancing the reaction equation, one can never obtain the correct answer! Then obtain the masses of octane reacted and carbon dioxide produced from the stoichiometric equation. After that, we now compare it with what is given as shown in the image attached.
To calculate the mass of carbon dioxide produced from the complete combustion of octane, you can use the balanced chemical equation and the molar masses of octane and CO2. First, calculate the number of moles of octane, then use the ratio of CO2 to octane to find the moles of CO2. Finally, multiply the moles of CO2 by the molar mass of CO2 to find the mass of CO2 produced.
Explanation:To calculate the mass of carbon dioxide produced from the complete combustion of octane (C8H18), you need to use the balanced chemical equation: 2 C8H18 + 25 O2 → 16 CO2 + 18 H2O. From this equation, you can see that for every 2 molecules of octane burned, 16 molecules of CO2 are generated. To find the mass of CO2 produced, you can use the molar mass of CO2 (44.01 g/mol) and the molar mass of octane (114.22 g/mol).
First, calculate the number of moles of octane: moles of octane = mass of octane / molar mass of octane = 16.6 g / 114.22 g/mol = 0.145 mol
Since 2 molecules of octane produce 16 molecules of CO2, you can multiply the number of moles of octane by the ratio of CO2 to octane: moles of CO2 = moles of octane x (16 mol CO2 / 2 mol octane) = 0.145 mol x (16 mol CO2 / 2 mol octane) = 1.16 mol CO2
Finally, calculate the mass of CO2 produced: mass of CO2 = moles of CO2 x molar mass of CO2 = 1.16 mol x 44.01 g/mol = 51.08 g
Therefore, the mass of carbon dioxide produced from the complete combustion of 16.6 g of octane is approximately 51.08 g.
Osmium has a density of 22.6 g/cm3. What volume (in cm3) would be occupied by a 21.8 g sample of osmium? 1.04 2.03 × 103 493 0.965 2.03 × 10-3
Answers
Answer:
volume =0.9645cm³
Explanation:
density refers to the ratio of mass to volume of a substance. It gives us a clue about how heavy an object is.
density = mass/volume
volume = mass/density
mass=21.8g
density=22.6g
volume =21.8/22.6
volume =0.9645cm³
The volume occupied by a 21.8 g sample of osmium with a density of 22.6 g/cm³ is approximately 0.9646 cm³, calculated by dividing the mass by the density.
To find the volume (in cm³) occupied by a 21.8 g sample of osmium, you can use the formula:
Volume = Mass / Density
Given:
- Mass (m) = 21.8 g
- Density (D) = 22.6 g/cm³
Now, plug these values into the formula:
Volume = 21.8 g / 22.6 g/cm³
Volume ≈ 0.9646 cm³
So, the volume occupied by the 21.8 g sample of osmium is approximately 0.9646 cm³.
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What is the mole fraction of benzene in the solution that has a vapor pressure of 38 torr at 20 ∘C? Express your answer using two significant figures.
Answers
The question is incomplete, here is the complete question:
At 20°C the vapor pressure of benzene [tex](C_6H_6)[/tex] is 75 torr, and that of toluene [tex](C_7H_8)[/tex] is 22 torr. Assume that benzene and toluene form an ideal solution.
What is the mole fraction of benzene in the solution that has a vapor pressure of 38 torr at 20°C? Express your answer using two significant figures.
Answer: The mole fraction of benzene is 0.302
Explanation:
Let the mole fraction of benzene be 'x' and that of toluene is '1-x'
To calculate the total pressure of the mixture of the gases, we use the equation given by Raoult's law, which is:
[tex]p_T=\sum_{i=1}^n(\chi_{i}\times p_i)[/tex]
We are given:
Vapor pressure of benzene = 75 torr
Vapor pressure of toluene = 22 torr
Vapor pressure of solution = 38 torr
Putting values in above equation, we get:
[tex]38=[(75\times x)+(22\times (1-x))]\\\\x=0.30[/tex]
Hence, the mole fraction of benzene is 0.30
Final answer:
The mole fraction of benzene in a solution with a vapor pressure of 38 torr at 20 C is calculated using Raoult's law with the vapor pressure of pure benzene. The calculation yields a value of approximately 0.5086, which is rounded to 0.51 when expressed in two significant figures.
Explanation:
Calculation of Mole Fraction of Benzene in a Solution
To calculate the mole fraction of benzene in a solution based on vapor pressure data, we use Raoult's law. According to Raoult's law, the vapor pressure of a solution component is equal to the product of the mole fraction of that component in the solution and the vapor pressure of the pure component. Since the question does not specify the total vapor pressure or the vapor pressure of pure benzene at 20°C, we will refer to the additional information provided that indicates at 20°C, the vapor pressures of pure benzene and toluene are 74.7 mmHg and 22.3 mmHg, respectively, to guide the calculation. Thus, the mole fraction (X) of benzene can be found by rearranging the Raoult's law equation: P = XP0, where P is the vapor pressure of the solution, and P0 is the vapor pressure of pure benzene. This rearrangement gives us X = P / P0.
Given that the solution has a vapor pressure of 38 torr (mmHg), and using the provided vapor pressure of pure benzene ( 74.7 mmHg), the calculation for the mole fraction of benzene (XC6H6) is: XC6H6 = 38 mmHg / 74.7 mmHg
From this calculation, the mole fraction of benzene is approximately 0.5086, which can be rounded to two significant figures to give a final answer of 0.51.
Phosphorus crystallizes in several different forms, one of which is a simple cube with an edge length of 238 pm. What is the density of phosphorus in the simple cubic crystal form?
Answers
Answer: The density of phosphorus is [tex]3.81g/cm^3[/tex]
Explanation:
To calculate the density of phosphorus, we use the equation:
[tex]\rho=\frac{Z\times M}{N_{A}\times a^{3}}[/tex]
where,
[tex]\rho[/tex] = density
Z = number of atom in unit cell = 1 (CCP)
M = atomic mass of phosphorus = 31 g/mol
[tex]N_{A}[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]
a = edge length of unit cell = [tex]238pm=238\times 10^{-10}cm[/tex] (Conversion factor: [tex]1cm=10^{10}pm[/tex] )
Putting values in above equation, we get:
[tex]\rho=\frac{1\times 31}{6.022\times 10^{23}\times (238\times 10^{-10})^3}\\\\\rho=3.81g/cm^3[/tex]
Hence, the density of phosphorus is [tex]3.81g/cm^3[/tex]