Answer:
p orbital.
Explanation:
Valence electrons are the electrons in an atom holding the very last orbital which is used in chemical bonding with other elements. Their existence could define the chemical properties of that atom.
During the first energy in ionization of an N2 molecule the molecular orbital from which the electron could be extracted is the only one with the highest energy level. Nitrogen has its outermost orbital (p) containing three valence electrons. Each orbital is only half filled, and thus it is unstable Thus, the electron mission must have been removed from p orbital.
For the first ionization energy for an N2 molecule, the molecular orbital that the electron is removed from is the p orbital.
It should be noted that valence electrons simply refer to the electrons in an atom that holds the last orbital that is required for chemical bonding with other elements.
The existence of valence electrons can define the chemical properties of that atom. For the first energy in ionization of an N2 molecule, the molecular orbital where the electron could be extracted is the p orbital since it has the highest energy level.
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If the metabolic rate of eggs at 25oC is 0.3 ml O2/hr and their metabolic rate at 35oC is 0.6 ml O2/hr, what is the Q10
Answer:
Q₁₀ = 2
Explanation:
The Q₁₀ can be calculated by the following equation:
[tex] Q_{10} = \frac{R_{2}}{R_{1}}^{10^{\circ} C/(T_{2}-T_{1})} [/tex]
where R: is the rate and T: is the temperature
With R₁=0.3 ml O₂/h, R₂=0.6 ml O₂/h, T₁=25 °C and T₂=35 °C, the Q₁₀ is:
[tex] Q_{10} = \frac{0.6 ml O_{2}/h}{0.3 ml O_{2}/h}^{10^{\circ} C/(35 -25)^{\circ} C} = 2 [/tex]
Therefore the Q₁₀ temperature coefficient of eggs is 2.
I hope it helps you!
A chemist measures the amount of bromine liquid produced during an experiment. He finds that 1.33 g of bromine liquid is produced. Calculate the number of moles of bromine liquid produced. Round your answer to 3 significant digits.
Answer:
8.32×10⁻³ moles of liquid Br₂ were produced.
Explanation:
You have the mass of bromine liquid produced and you must calculate the moles. Therefore you divide mass / molar mass Br
Bromine is a diatomic molecule, Br₂ so molar mass is 159.8 g/mol
1.33 g / 159.8 g/mol = 8.32×10⁻³ moles
Which one of the following substances would be the least soluble in C6H12?
(a.) I2
(b.) Na2SO4
(c.) HI
(d.) C2H6
(e.) CH3CH2OH
Answer: The least soluble will be Na2SO4
Explanation:
C6H12, is the chemical formula for cyclohexane, a ORGANIC compound and volatile liquid.
Recall that, organic solvents do dissolve organic solutes. Now, of all the substances given in the options, NA2SO4 is the most ionic and inorganic as well.
Thus, it will be the least soluble
The substance which would be least soluble in C6H12 is; Choice B; Na2SO4
Solubility of Organic compounds
By convention, organic compounds are veryich soluble in organic compounds.
Hence, compounds C2H6 and CH3CH2OH would be very much soluble in the compound, C6H12.
The compound which is least soluble in C6H12 is the inorganic and highly ionic Na2SO4.
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What volume of 3.00 M HCl will form a solution with an acidic pH when mixed with 100 mL of 3.00 M NaOH
Answer: The volume of acid should be less than 100 mL for a solution to have acidic pH
Explanation:
To calculate the volume of acid needed to neutralize, we use the equation given by neutralization reaction:
[tex]n_1M_1V_1=n_2M_2V_2[/tex]
where,
[tex]n_1,M_1\text{ and }V_1[/tex] are the n-factor, molarity and volume of acid which is HCl
[tex]n_2,M_2\text{ and }V_2[/tex] are the n-factor, molarity and volume of base which is NaOH
We are given:
[tex]n_1=1\\M_1=3.00M\\V_1=?mL\\n_2=1\\M_2=3.00M\\V_2=100mL[/tex]
Putting values in above equation, we get:
[tex]1\times 3.00\times V_1=1\times 3.00\times 100\\\\V_1=\frac{1\times 3.00\times 100}{1\times 3.00}=100mL[/tex]
For a solution to be acidic in nature, the pH should be less than the volume of acid needed to neutralize.
Hence, the volume of acid should be less than 100 mL for a solution to have acidic pH
The half-lives for the forward and reverse reactions that are first order in both directions are 24ms and 39 ms, respectively. Calculate the corresponding relaxation time (s) for the return to equilibrium after a temperature jump. Please enter your answer with two significant figures (unit: s).
Explanation:
The given data is as follows.
[tex](t_{\frac{1}{2}})_{1}[/tex] = 24 ms,
[tex](t_{\frac{1}{2}})_{2}[/tex] = 39 ms,
where, [tex](t_{\frac{1}{2}})_{1}[/tex] = half-life for the forward reaction
[tex](t_{\frac{1}{2}})_{2}[/tex] = half-life for the backward reaction
It is known that the formula for first-order reaction is as follows.
[tex]t_{\frac{1}{2}} = \frac{0.693}{K}[/tex]
Therefore,
[tex]K_{1} = \frac{0.693}{(t_{\frac{1}{2}})_{1}}[/tex]
= [tex]\frac{0.693}{24 ms}[/tex]
= 0.0289 [tex]ms^{-1}[/tex]
[tex]K_{2} = \frac{0.693}{(t_{\frac{1}{2}})_{2}}[/tex]
= [tex]\frac{0.693}{39 ms}[/tex]
= 0.0178 [tex]ms^{-1}[/tex]
Hence, formula for the relaxation time is as follows.
[tex]\tau = \frac{1}{K_{1} + k_{2}}[/tex]
= [tex]\frac{1}{(0.0289 + 0.0178) ms^{-1}}[/tex]
= 21.41 ms
Thus, we can conclude that the corresponding relaxation time(s) for the return to equilibrium after a temperature jump is 21.41 ms.
The resolution limit of a microscope is roughly equal to the wavelength of light used in producing the image. Electron microscopes use an electron beam (in place of photons) to produce much higher resolution images, about 0.22 nm in modern instruments.Assuming that the resolution of an electron microscope is equal to the de Broglie wavelength of the electrons used, to what speed must the electrons be accelerated to obtain a resolution of 0.26 ?
Explanation:
Equation for de Broglie wavelength is as follows.
[tex]\lambda = \frac{h}{mv}[/tex]
where, m = mass of particle moving
v = velocity
h = Planck's constant = [tex]6.626 \times 10^{-34} Js[/tex]
When particle is electron then value of mass is [tex]9.109 \times 10^{-31} kg[/tex].
Wavelength of electron = 0.26 nm
= [tex]0.26 nm \times \frac{10^{-9}}{1 nm}[/tex]
= [tex]0.26 \times 10^{-9}[/tex] nm
Therefore, speed of electron will be calculated as follows.
v = [tex]\frac{h}{m \lambda}[/tex]
= [tex]\frac{6.626 \times 10^{-34} Js}{2.6 \times 10^{-10} \times 9.109 \times 10^{-31} kg}[/tex]
= [tex]2.79 \times 10^{6} m/s[/tex]
Thus, we can conclude that speed at which electrons be accelerated to obtain a resolution of 0.26 is [tex]2.79 \times 10^{6} m/s[/tex].
To obtain a resolution of 0.26 nm in an electron microscope, electrons need to be accelerated to a speed of approximately 2.75 × 10⁶ m/s. This is based on the de Broglie equation which relates the speed of a particle to its wavelength.
Explanation:The question is about determining the speed that electrons need to be accelerated to in an electron microscope so as to achieve a resolution of 0.26 nm. This can be determined using the de Broglie equation, which relates the speed of a particle to its wavelength. The equation is λ = h/mv, where h (Planck's constant) is 6.626 × 10⁻³⁴ m² kg / s, m (the mass of an electron) is 9.11 × 10⁻³¹ kg, and v is the velocity or speed of the electron.
To find v, we can rearrange the equation: v = h/mλ. If we substitute the given values (λ = 0.26 nm or 0.26 × 10⁻⁹ m), we get:
v = (6.626 × 10⁻³⁴ m² kg / s) / (9.11 × 10⁻³¹ kg × 0.26 × 10⁻⁹ m) ≈ 2.75 × 10⁶ m/s.
Therefore, the electrons need to be accelerated to a speed of approximately 2.75 × 10⁶ m/s to achieve a resolution of 0.26 nm in an electron microscope.
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g Tibet (altitude above sea level is 29,028 ft) has an atmospheric pressure of 240. mm Hg. Calculate the boiling point of water in Tibet. The heat of vaporization for water is 40.7 kJ/mol.
Answer: The boiling point of water in Tibet is 69.9°C
Explanation:
To calculate the boiling point of water in Tibet, we use the Clausius-Clayperon equation, which is:
[tex]\ln(\frac{P_2}{P_1})=\frac{\Delta H}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]P_1[/tex] = initial pressure which is the pressure at normal boiling point = 1 atm = 760 mmHg (Conversion factor: 1 atm = 760 mmHg)
[tex]P_2[/tex] = final pressure = 240. mmHg
[tex]\Delta H_{vap}[/tex] = Heat of vaporization = 40.7 kJ/mol = 40700 J/mol (Conversion factor: 1 kJ = 1000 J)
R = Gas constant = 8.314 J/mol K
[tex]T_1[/tex] = initial temperature or normal boiling point of water = [tex]100^oC=[100+273]K=373K[/tex]
[tex]T_2[/tex] = final temperature = ?
Putting values in above equation, we get:
[tex]\ln(\frac{240}{760})=\frac{40700J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{T_2}]\\\\-1.153=4895.36[\frac{T_2-373}{373T_2}]\\\\T_2=342.9K[/tex]
Converting the temperature from kelvins to degree Celsius, by using the conversion factor:
[tex]T(K)=T(^oC)+273[/tex]
[tex]342.9=T(^oC)+273\\T(^oC)=(342.9-273)=69.9^oC[/tex]
Hence, the boiling point of water in Tibet is 69.9°C
Draw the structures of the starting materials needed to make 2-methylhept-3-yne in the spaces provided. The starting materials may be any bromoalkane having five carbons or fewer.
Answer: answered
Explanation:
A military jet cruising at an altitude of 12.0kmand speed of 1900./kmh burns fuel at the rate of 74.4/Lmin. How would you calculate the amount of fuel the jet consumes on a 1050.km mission? Set the math up. But don't do any of it. Just leave your answer as a math expression.
Answer: amount = 2466.95L
Explanation:
given that the speed is = 1900./kmh i.e. 1hr/900km
distance = 1050km
the fuel burns at a rate of 74.4 L/min
therefore the amount of fuel that the jet consumes on a 1050.km becomes;
total fuel used = time × fuel burning rate
where time = distance / speed
∴ total fuel used (consumed) = time × fuel burning rate
total fuel consumed = (1050km × 1hr/1900km) × (60min/ 1hr × 74.4L/1min)
total fuel consumed = 2466.95L
Where would you expect to find the 1H1H NMR signal of (CH3)2Mg(CH3)2Mg relative to the TMS signal
Answer:
In diethyl ether, (CH₃)₂Mg(CH₃)₂Mg, show peaks at -1.46 and -1.74 ppm, at -100 °CIn tetrahydrofuran (THF), monomer (CH₃)₂Mg show peaks at -1.76 and -1.81 ppm, at -60 °CExplanation:
The 1H NMR signal of (CH₃)₂Mg can be obtained in diethyl ether and tetrahydrofuran. (CH₃)₂Mg tends to polymerize in diethyl ether but mostly remain monomer in tetrahydrofuran. The negative value of chemical shifts shows that the protons of (CH₃)₂Mg are more shielded than the protons of TMS.
In diethyl ether, the structure of (CH₃)₂Mg is predicted in Figure A (attached). The A group in the structure represents the solvent. The chemical shift at -1.46 ppm is associated with protons of the bridging methyl groups while the signal at -1.74 ppm is for the proton of the terminal methyl group.
In THF, the two possible structures of (CH₃)₂Mg are predicted in Figure B (attached). Chemical shift at -1.76 ppm is associated with unsolvated monomer of (CH₃)₂Mg while the shift at -1.81 refers to the protons of methyl group bonded to the solvated (CH₃)₂Mg. The solvent is represented as Y in the Figure B.
This study is reported by J. Heard in "NMR of organomagnessium compounds".
A lithium atom starts from rest at a position x = xo and falls toward zinc atom with an acceleration that depends on their separation as a(x) = 4Eϵ/m { 12 (σ^12/σ^13) -6 (σ^6/σ^7) where ϵ, m, and σ are positive constants, and the zinc atom stays fixed at position x = 0, Assume that xo > 2^(1/6)σ. a. Find an expression for how the velocity of the moving lithium atom depends on x. (It will also depend on ϵ, m, xo and σ) b. What is the distance of closest approach of the the two atoms (in terms of σ and xo)?
Answer:
Please refer to the attachment for answers.
Explanation:
Please refer to the attachment for explanation
At 0 °C and 1.00 atm, as much as 0.70 g of O2 can dissolve in 1 L of water. At 0 °C and 4.00 atm, how many grams of O2 dissolve in 1 L of water?
Answer : The mass of [tex]O_2[/tex] at 4.00 atm pressure is, 2.8 grams.
Explanation :
First we have to calculate the Henry law constant.
As we know that,
[tex]C_{O_2}=k_H\times p_{O_2}[/tex]
where,
[tex]C_{O_2}[/tex] = solubility of [tex]O_2[/tex] = 0.70 g/L
[tex]p_{O_2}[/tex] = partial pressure of [tex]O_2[/tex] = 1.00 atm
[tex]k_H[/tex] = Henry's law constant
Now put all the given values in the above formula, we get:
[tex]0.70g/L=k_H\times (1.00atm)[/tex]
[tex]k_H=0.70g/L.atm[/tex]
Now we have to calculate the solubility of [tex]O_2[/tex] at 4.00 atm pressure.
[tex]C_{O_2}=k_H\times p_{O_2}[/tex]
where,
[tex]C_{O_2}[/tex] = solubility of [tex]O_2[/tex] at 4.00 atm = ?
[tex]p_{O_2}[/tex] = partial pressure of [tex]O_2[/tex] = 4.00 atm
[tex]k_H[/tex] = Henry's law constant = [tex]0.70g/L.atm[/tex]
Now put all the given values in the above formula, we get:
[tex]C_{O_2}=(0.70g/L.atm)\times (4.00atm)[/tex]
[tex]C_{O_2}=2.8g/L[/tex]
Now we have to calculate the mass of [tex]O_2[/tex]
[tex]\text{Mass of }O_2=\text{Solubility of }O_2\times \text{Volume of solution}[/tex]
[tex]\text{Mass of }O_2=2.8g/L\times 1L=2.8g[/tex]
Thus, the mass of [tex]O_2[/tex] at 4.00 atm pressure is, 2.8 grams.
The amount of O2 that can dissolve in 1L of water at 0 °C and 4.00 atm, using Henry's law, is 2.80 g.
Explanation:In this question, we determine the amount of O2 that can dissolve in water at a higher pressure using Henry's law.
According to Henry's law, the amount of dissolved gas in a liquid is directly proportional to its partial pressure above the liquid. If the pressure is increased to 4 atm (4 times the initial pressure), the amount of O2 that can dissolve in the water will also increase 4 times.
Therefore, at 0 °C and 4.00 atm, 4 * 0.70 g = 2.80 g of O2 can dissolve in 1L of water.
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What mass of oxygen gas is produced by the reaction of 3.8 g of ammonium perchlorate?
Answer:
There is 1.29 grams of O2 produced.
Explanation:
Step 1: Data given
Mass of ammonium perchlorate 3.8 grams
Molar mass of ammonium perchlorate = 117.49 g/mol
Step 2: The balanced equation
4NH4ClO4 → 5O2 + 6H2O + 2N2 + 4HCl
Step 3: Calculate moles of NH4ClO4
Moles NH4ClO4 = mass NH4ClO4 / molar mass NH4ClO4
Moles NH4ClO4 = 3.80 grams / 117.49 g/mol
Moles NH4ClO4 = 0.0323 moles
Step 4: Calculate moles of O2
For 4 moles of ammonium perchlorate consumed, we produce 5 moles O2, 6 moles H2O, 2 moles N2 and 4 moles HCl
For 0.0323 moles NH4ClO4 we'll have 5/4 * 0.0323 = 0.040375 moles of O2
Step 5: Calculate mass of O2
Mass O2 = moles O2 * molar mass O2
Mass O2 = 0.040375 moles * 32 g/mol
Mass O2 = 1.29 grams
There is 1.29 grams of O2 produced.
The mass of oxygen gas produced by the reaction of 3.8 g g ammonium perchlorate is 1.03 grams.
The balanced equation for the reaction of ammonium perchlorate is:
NH₄ClO₄ → 2H₂O + 2HCl + O₂
For every 1 mole of ammonium perchlorate, 1 mole of oxygen gas (O₂) is produced.
The molar mass of ammonium perchlorate (NH₄ClO₄) is calculated as follows:
NH₄ClO₄: (1 × 14.01 g/mol) + (4 × 1.01 g/mol) + (1×* 35.45 g/mol) + (4 × 16.00 g/mol) = 117.49 g/mol
Moles of NH₄ClO₄ = mass / molar mass = 3.8 g / 117.49 g/mol
For every mole of ammonium perchlorate, 1 mole of oxygen gas is produced. Therefore, the number of moles of oxygen gas produced is equal to the number of moles of ammonium perchlorate.
Moles of O₂ = Moles of NH₄ClO₄
Mass of O₂ = moles of O₂ × molar mass of O₂
Mass of O₂ = moles of O₂ × molar mass of O₂ = moles of NH₄ClO₄ × molar mass of O₂
Mass of O₂ = (3.8 g / 117.49 g/mol) × 32.00 g/mol = 1.03 g
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Deprenyl is an enzyme inhibitor that helps prevent the metabolism of dopamine in the brain. The chemical formula of Deprenyl is C13H17NHCl. The appropriate dose for treating Parkinson’s disease is 100 μg/(day*kg body weight). For a 70 kg
Answer:
0.007 g of deprenyl dose is required fro the patient with body mass of 70 kilograms.
Explanation:
The dose for treating Parkinson’s disease = 100 μg/kg body weight
Mass of patient's body = 70 kg
Amount of dose of deprenyl required = 100 μg/kg × 70 kg = 7,000 μg
1 μg = 0.00001 g
7,000 μg = 7,000 × 0.000001 g = 0.007 g
0.007 g of deprenyl dose is required fro the patient with body mass of 70 kilograms.
Deprenyl is a chemical compound used as a treatment for Parkinson's disease. It works by inhibiting the enzyme that breaks down dopamine, increasing dopamine levels in the brain. The appropriate dose for treating Parkinson's disease with Deprenyl is 100 μg/day per kg of body weight.
Explanation:Deprenyl is a chemical compound with the formula C13H17NHCl. It is an enzyme inhibitor that helps prevent the metabolism of dopamine in the brain. Deprenyl is commonly used as a treatment for Parkinson's disease, a neurodegenerative disorder characterized by a loss of dopaminergic neurons.
Deprenyl works by inhibiting the enzyme monoamine oxidase-B (MAO-B), which is responsible for breaking down dopamine in the brain. By preventing the breakdown of dopamine, Deprenyl helps to increase dopamine levels, which can alleviate symptoms of Parkinson's disease.
The appropriate dose of Deprenyl for treating Parkinson's disease is 100 μg/day per kg of body weight. For example, for a 70 kg individual, the appropriate dose would be 7000 μg or 7 mg per day.
Complete the mechanism for the acid-catalyzed hydrolysis of the epoxide in alcohol by adding any missing atoms, bonds, charges, nonbonding electrons, and curved arrows.Figure:Different bonds in the figure
The acid-catalyzed hydrolysis of epoxides in alcohol involves the step-wise transfer of protons and the breaking and formation of various bonds. Large quantities of ethanol can be synthesized through addition reactions, which also creates alcohols - organic compounds with a hydroxyl group bonded to a carbon atom.
Explanation:The acid-catalyzed hydrolysis of epoxides in alcohol is a process that may involve the sequential or step-wise transfer of protons, aided by the conjugate base and bonded water molecules acting as acids. Large quantities of ethanol can be synthesized through the addition reaction of water with ethylene using an acid as a catalyst, which is indicative of the importance of these actions in chemical reactions. Notably, this also involves the breaking and forming of various bonds, such as breaking a C-O triple bond and two H-H single bonds, alongside forming three C-H single bonds, a C-O single bond, and an O-H single bond.
For clarification, an addition reaction is a typical alkene reaction where a double carbon-carbon bond forms a single carbon-carbon bond by adding a reactant. Originating from this process are alcohols, organic compounds consisting of a hydroxyl group (-OH) bonded to a carbon atom.
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A lead ball has a mass of 55.0 grams and a density of 11.4 g/cm3. What is the volume of the ball?
A) 0.207 mL
B) 0.207 L
C) 4.82 mL
D) 4.82 L
E) none of the above
Answer:
B
Explanation:
Answer:
The volume of the ball is 4.82 mL (opton C)
Explanation:
Step 1: Data given
Mass of the lead ball = 55.0 grams
Density = 11.4 g/cm³
Step 2: Calculate volume of the balloon
Volume = mass / density
Volume = 55.0 g / 11.4 g/cm³
Volume = 4.82 cm³ = 4.82 mL
The volume of the ball is 4.82 mL (opton C)
In the molecular orbital model of benzene, the six p atomic orbitals combine to form how many molecular orbitals?
Answer:
6 molecular orbital.
Explanation:
According to the molecular orbital theory, number of molecular orbital are obtained is equal to the number of atomic orbital combine.
When six p orbitals of benzene combine, the will form six molecular orbital. These six molecular orbital have 3 molecular orbital with lower energy and three molecular orbital with higher energy.
The orbital with lower energies are called bonding molecular orbital and orbital with higher energies are called antibonding molecular orbital.
In the molecular orbital model of benzene, six p atomic orbitals combine to form six molecular orbitals. These result from the overlapping p orbitals of the six carbon atoms in the benzene ring, forming a 'pi electron cloud'.
Explanation:In the molecular orbital model of benzene, the six p atomic orbitals combine to form six molecular orbitals. This happens because benzene has a cyclic structure where the overlapping p orbitals of six carbon atoms form what is known as a 'pi electron cloud'. This pi electron cloud is a result of side-on overlap of p orbitals encompassing all the six carbon atoms. So, essentially, the six overlapping p orbitals produce six molecular orbitals. These six orbitals then distribute themselves into differing energy levels. Three of these orbitals tend to be lower energy (bonding), and the other three are higher energy (anti-bonding).
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A reaction in which a substance reacts with oxygen, emitting heat and forming oxygen-containing compounds is an example of a(n):
A) acid-base reaction.
B) combustion reaction.
C) precipitation reaction.
D) gas evolution reaction.
Answer: B) combustion reaction.
Explanation:
A) acid-base reaction: When an acid reacts with a base, to form metal salt and water, this type of reaction is Acid Base reaction.
Example: [tex]HCl+NaOH\rightarrow NaCl+H_2O[/tex]
B) combustion reaction: When a hydrocarbon reacts with oxygen to produce carbon dioxide and water, this type of reaction is combustion reaction.
[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]
C) precipitation reaction: a reaction in which aqueous solution of two compounds on mixing react to form an insoluble compound which separate out as a solid are called precipitation reactions.
[tex]Na_2SO_4(aq)+BaCl_2(aq)\rightarrow BaSO_4(s)+NaCl(aq)[/tex]
D) gas evolution reaction: a reaction in which one of the product is formed as a gas.
[tex]CaCO_3(s)\rightarrow CaO(s)+CO_2(g)[/tex]
Calculate the volume in milliliters of a 1.72 mol/L sodium thiosulfate solution that contains 375. mmol of sodium thiosulfate (Na_2S_2O_3).
Answer:
216 mL
Explanation:
Molarity of a substance , is the number of moles present in a liter of solution .
M = n / V
M = molarity ( unit = mol / L or M )
V = volume of solution in liter ( unit = L ),
n = moles of solute ( unit = mol ),
From the question , the respective values are as follows -
M = 1.73 mol / L
n = 375 mmol
Since,
1 mmol = 0.001 mol
n = 375 * 0.001 mol = 0.375 mol
Now , the volume of can be calculated by using the above equation ,
M = n / V
Putting the respective values ,
1.73 mol / L = 0.375 mol / V
V = 0.216 L
Since,
1 L = 1000 mL
V = 0.216 * 1000 mL = 216 mL
A 75.0 g piece of gold at 650. K is dropped into 180. g of H2O(l) at 310. K in an insulated container at 1 bar pres- sure. Calculate the temperature of the system once equilib- rium has been reached. Assume that CP, m for Au and H2O is constant at their values for 298 K throughout the temperature range of interest.
Answer:
The temperature of the system after reaching equilibrium is 314.21 K.
Explanation:
Heat lost by gold will be equal to heat gained by the water
[tex]-Q_1=Q_2[/tex]
Mass of iron = [tex]m_1=75.0 g[/tex]
Specific heat capacity of gold= [tex]c_1=0.126 J/gK [/tex]
Initial temperature of the gold= [tex]T_1=650.0 K [/tex]
Final temperature of gold = [tex]T_2[/tex]=T
[tex]Q_1=m_1c_1\times (T-T_1)[/tex]
Mass of water= [tex]m_2=180.0 g[/tex]
Specific heat capacity of water= [tex]c_2=4.184 J/gK [/tex]
Initial temperature of the water = [tex]T_3=310 K[/tex]
Final temperature of water = [tex]T_2[/tex]=T
[tex]Q_2=m_2c_2\times (T-T_3)[/tex]
[tex]-Q_1=Q_2[/tex]
[tex]-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)[/tex]
On substituting all values:
[tex]-(75.0 g\times 0.126 J/gK\times (T-650.0K))=180.0 g\times 4.184 J/gK\times (T-310.0K)[/tex]
we get, T = 314.21 K
The temperature of the system after reaching equilibrium is 314.21 K.
Final answer:
The question asks for the final temperature of a gold and water system at thermal equilibrium, a concept from chemistry involving heat transfer calculations.
Explanation:
The question involves calculating the final temperature of a system consisting of gold and water once they reach thermal equilibrium. This is a classic problem in thermodynamics, specifically within the scope of heat transfer and calorimetry. According to the principle of conservation of energy, the heat lost by the gold will be equal to the heat gained by the water until equilibrium is reached. However, the exact calculation requires knowing the specific heat capacities of gold and water, which are not provided in the query. Typically, you would use the formula q=mcΔT (where q is the heat absorbed or released, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature) for both substances and set the heat lost by gold equal to the heat gained by water to find the final equilibrium temperature.
Draw relative enthalpy diagrams showing the relationship between the enthalpy ofhydration, lattice enthalpy, and the solvation enthalpy of solution for KI, CaCl2andNaCl dissolving in water.
Complete Question:
The complete question is on the first uploaded image
Answer:
The solution is on the second uploaded image
Explanation:
The explanation of the above solution is on the third uploaded image.
The second-order diffraction for a gold crystal is at an angle of 22.20o for X-rays of 154 pm. What is the spacing between the crystal planes?
Answer: The spacing between the crystal planes is [tex]4.07\times 10^{-10}m[/tex]
Explanation:
To calculate the spacing between the crystal planes, we use the equation given by Bragg, which is:
[tex]n\lambda =2d\sin \theta[/tex]
where,
n = order of diffraction = 2
[tex]\lambda[/tex] = wavelength of the light = [tex]154pm=1.54\times 10^{-10}m[/tex] (Conversion factor: [tex]1m=10^{12}pm[/tex] )
d = spacing between the crystal planes = ?
[tex]\theta[/tex] = angle of diffraction = 22.20°
Putting values in above equation, we get:
[tex]2\times 1.54\times 10^{-10}=2d\sin (22.20)\\\\d=\frac{2\times 1.54\times 10^{-10}}{2\times \sin (22.20)}\\\\d=4.07\times 10^{-10}m[/tex]
Hence, the spacing between the crystal planes is [tex]4.07\times 10^{-10}m[/tex]
Final answer:
The spacing between the crystal planes can be found using Bragg's law, and using the given values for a second-order diffraction and the X-ray wavelength, the calculation will yield the required plane spacing.
Explanation:
The question is about determining the spacing between the crystal planes using X-ray diffraction data. When X-rays are diffracted by the crystal planes, the relationship between the angle of diffraction (Bragg angle), the wavelength of the X-rays, and the spacing between the planes is given by Bragg's law:
Bragg's Law: nλ = 2d sin(θ)
Where:
n is the order of diffractionλ (lambda) is the wavelength of the X-raysd is the distance between the crystal planesθ (theta) is the Bragg angleFor the given problem where a second-order diffraction for a gold crystal is observed at an angle of 22.20° for X-rays of 154 pm (0.154 nm), we can rearrange Bragg's law to solve for d:
d = nλ / (2 sin(θ))
Using the provided values, we can calculate:
d = 2(0.154 nm) / (2 sin(22.20°))
This calculation yields the spacing between the crystal planes, which is the answer to the student's question.
What is the frequency of green light that has a wavelength of 543 nm? (c = 3.00 x 10⁸ m/s)
Answer:
ν = 5.53 x 10⁶ Hz
Explanation:
The frequency of a radiation, ν ,is inversely propotional to its wavelength, and is given by:
ν = c/λ where
ν = frequency
c: speed of light, 3 x 10⁸ m/s
λ: wavelength in m
Think of frequency as the number of waves per second.
We can directly compute the answer by first converting the wavelength to m for unit consistency and then plugging the values.
1 m = 10⁹ nm
543 nm x ( 1 m / 10⁹ nm ) = 5.43 x 10⁻⁷ m
ν = c/λ = 3 x 10⁸ m/s / 5.43 x 10⁻⁷ m = 5.53x 10¹⁴ s⁻¹
In the metric system the unit s⁻¹ is called Hertz.
ν = 5.53 x 10¹⁴ Hz
The frequency of green light with a wavelength of 543 nm is approximately 5.52 x 10^14 Hz.
Explanation:The frequency of electromagnetic waves can be calculated using the equation: frequency = speed of light / wavelength. In this case, we have a wavelength of 543 nm and the speed of light is 3.00 x 10^8 m/s. First, we need to convert the wavelength to meters. There are 1 x 10^9 nm in a meter, so the wavelength is 543 x 10^-9 m. Plugging these values into the equation, the frequency of the green light is approximately 5.52 x 10^14 Hz.
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A goose is flying south for the winter at a constant speed. Keep in mind that one mile is 1.61 km, and one pound is 454 g.The goose has a mass of 20.1 lb and is flying at 9.60 miles/h. What is the kinetic energy of the goose in joules?
Answer:
kinetic energy = 84.0260 joules
Explanation:
given data
one mile = 1.61 km
one pound = 454 g
mass = 20.1 lb = 9.1172 kg
flying velocity = 9.60 miles/h = 9.60 × 1.61 × [tex]\frac{5}{18}[/tex] = 4.2933 m/s
solution
we get here kinetic energy that is express as
kinetic energy = 0.5 × m × v² .........................1
put here value and we get
kinetic energy = 0.5 × 9.1172 × 4.2933²
kinetic energy = 84.0260 joules
The kinetic energy of the goose, when converted to the same units, is approximately 83.74 joules.
Explanation:The kinetic energy for any object can be found using the equation K.E. = 1/2*m*v^2, where m is mass and v is the velocity. First we need to convert the mass of the goose from pounds to kilograms, and the speed from miles per hour to meters per second. So, the mass should be 20.1 lb * 454 g = 9121.4 g. Converting that to kilograms, we get 9121.4 g * 1 kg/1000 g = 9.1214 kg. The speed of the goose should be 9.6 miles/hour * 1.61 km = 15.44 km/h. Converting that to meters per second, we get 15.44 km/h * 1000 m/1 km * 1 h/3600 s = 4.2889 m/s. Finally, we can substitute these values into the kinetic energy equation
K.E = 1/2 * 9.1214 kg * (4.2889 m/s)^2 = 83.74 joules.
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IN TEH HYPOTHETIcal case of aliquid that had such strong intermolecular forces it would never evaporate, what temperature change would you expect to see
Answer:
A wide temperature change.
Explanation:
The intermolecular force is the force that puts the molecules together in a substance. If the substance is ionic (formed by ions) than the force is called ion-ion, which is very strong. If it's a covalent compound (the elements share electron pairs), then it can be a polar or a nonpolar substance.
The polar substance has a huge difference in electronegativity of its components, so partial charges are presented. In this case, the intermolecular force is called dipole-dipole, because charged poles are formed. In the other case, of nonpolar, the dipoles are induced, so it's called dipole induced-dipole induced force.
When a polar compound has a hydrogen-bonded to a high electronegative element (F, O, or N), the dipole-dipole is extremely strong and it's called a hydrogen bond.
Thus, as strong is the force (Ionic > hydrogen bond > dipole-dipole > dipole induced - dipole induced), as difficult is to broken these bonds. In a phase change, those bonds must be broken. So if the liquid has so strong forces, it would be necessary a large amount of energy to evaporate it.
The temperature is a measure of the kinetic energy of the molecules, so if the energy applied is too high, the temperature change must be extremely high too!
This situation is not real, because if the forces are so strong than the material would be a solid and not a liquid. Besides, even an extremely strong bond may be broken with the right temperature increase, so the liquid would evaporate!
In a hypothetical liquid with intermolecular forces strong enough to prevent evaporation, we would not expect to see a temperature change due to evaporation. The liquid's temperature would be unaffected by this specific phase change process and would only change due to external thermal influences. This is a theoretical extreme, as in reality, all substances can change phase with sufficient energy.
Explanation:Temperature Change in a Substance with Strong Intermolecular Forces
If we hypothesize a liquid with such strong intermolecular forces that it would never evaporate, we must consider the concept of vapor pressure. Typically, vapor pressure represents the tendency of molecules to escape into the vapor phase; stronger intermolecular forces would mean a lower vapor pressure. In this hypothetical scenario, because evaporation happens when molecules have enough kinetic energy to overcome intermolecular forces, and these forces are too strong to be overcome, one would not expect to see any temperature change due to evaporation.
Moreover, since evaporation is an endothermic process where the liquid absorbs energy to allow molecules to escape, the absence of evaporation implies that the liquid would not absorb energy in this way. Therefore, the temperature of the liquid would remain relatively stable, primarily affected only by external thermal inputs or losses, but not through phase change.
It is important to realize that this is an extreme hypothetical situation, as all real substances have a point at which their intermolecular bonds can be overcome by kinetic energy, leading to a phase transition. Additionally, in real-world scenarios, temperature impacts the kinetic energy of molecules within a substance, potentially affecting state changes at varying temperatures.
Explain the reason potassium was visible when using the cobalt glass. Describe what occured.
Explanation:
If potassium is burnt the ions go into a high state of energy. Once they cool, it gives off energy in the form of a visible spectrum which has a characteristic color Now, The cobalt glass blocks out yellow light, and potassium ion which is purple in color is visible.
Final answer:
Cobalt glass filters out other wavelengths of light, allowing the violet flame of potassium to be seen clearly during a flame test, thereby facilitating the identification of potassium by its characteristic color.
Explanation:
Potassium was visible using cobalt glass because cobalt glass has the ability to filter out certain wavelengths of light and allows only specific wavelengths to pass through. This element-specific coloration aids the identification of potassium because potassium compounds, when heated, emit a violet or lilac color that can be masked by the colors of other substances. However, the cobalt glass helps by primarily allowing those violet wavelengths to pass through, making it easier to observe the potassium flame test result clearly.
The illustration of the electron transfer to form potassium sulfide (K₂S) would involve two potassium atoms each donating one electron to a sulfur atom, with each potassium atom becoming a K⁺ ion, and the sulfur atom gaining two electrons to become an S²⁻ ion.
A metal ion with a net +3 charge has five electrons in the 3d shell. Identify the metal.
Answer: The given metal is iron.
Explanation:
We are given:
The electrons in 3d-shell = 5
When the metal is present in a transition series (here it is present in first transition series), the removal of valence electron takes place from '4s' shell first and then from '3d' shell
Electronic configuration is defined as the representation of electrons around the nucleus of an atom.
Number of electrons in an atom is determined by the atomic number of that atom.
Inner shell electrons for the given element = 18
Total number of electrons in the ion = 18 + 5 = 23
We are given:
Charge on the ion = +3
To calculate the number of electrons, we use the equation:
Number of electrons = Atomic number - charge
Atomic number = 23 + 3 = 26
The element having atomic number '26' is iron
Hence, the given metal is iron.
Answer: the metal is Iron (Fe)
Explanation: The original electronic configuration of Fe is 1s2 2s2 2p6 3s2 3p6 3d6 4s2
Because it has no charge. The electronic configuration will change from the original because the metal has a charge of +3, Fe is element number 26 on the periodic table b ut with the +3 charge, the element number will change to 23 i.e 26-(+3) =23
The configuration for Fe3+ is 1s2 2s2 2p6 3s2 3p6 3d5, it automatically had 5 electron in 3d shell
What is the concentration of magnesium bromide, in ppm, if 133.4 g MgBr2 dissolved in 1.84 L water. Then solve for the bromine concentration in ppm
Answer: 0.0725ppm
Explanation:
133.4g of MgBr2 dissolves in 1.84L of water.
Therefore Xg of MgBr2 will dissolve in 1L of water. i.e
Xg of MgBr2 = 133.4/1.84 = 72.5g
The concentration of MgBr2 is 72.5g/L = 0.0725mg/L
Recall,
1mg/L = 1ppm
Therefore, 0.0725mg/L = 0.0725ppm
The concentration of magnesium bromide (MgBr2) in the solution is 72467.39 ppm. Considering each molecule of MgBr2 has two atoms of bromine, the concentration of bromine in the solution is 58032.60 ppm.
Explanation:The subject of this question is Chemistry, specifically dealing with the concept of concentration, which is a measure of the amount of solute dissolved in a solvent. The units of ppm (parts per million) imply a measurement of the amount of a particular substance (in this case magnesium bromide and bromine) in a total amount of 1 million parts of the mixture.
First, we find the concentration of magnesium bromide MgBr2 by using the formula ppm = (mass of solute/volume of solution) x 10^6. Given that we have 133.4 g of MgBr2 in 1.84 L of water, we find that the concentration of MgBr2 in ppm is (133.4 g/1.84 L) x 10^6 = 72467.39 ppm.
Next, to solve for the concentration of bromine in ppm, we must consider that each molecule of MgBr2 has two atoms of Br. So, the mass of bromine in 133.4 g of MgBr2 is 133.4 g * (79.9 g/mol Br /159.8 g/mol MgBr2) * 2 mol Br = 106.76 g. The concentration of bromine in ppm is then (106.76 g / 1.84 L) x 10^6 = 58032.60 ppm.
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Calculate the acid dissociation constant Ka of a 0.2 M solution of weak acid that is 0.1% ionized is ________.
Answer: acid dissociation constant Ka= 2.00×10^-7
Explanation:
For the reaction
HA + H20. ----> H3O+ A-
Initially: C. 0. 0
After : C-Cx. Cx. Cx
Ka= [H3O+][A-]/[HA]
Ka= Cx × Cx/C-Cx
Ka= C²X²/C(1-x)
Ka= Cx²/1-x
Where x is degree of dissociation = 0.1% = 0.001 and c is the concentration =0.2
Ka= 0.2(0.001²)/(1-0.001)
Ka= 2.00×10^-7
Therefore the dissociation constant is
2.00×10^-7
The second-order rate constant for the following gas-phase reaction is 0.041 1/MLaTeX: \cdotâs. We start with 0.438 mol C2F4 in a 2.42 liter container, with no C4F8 initially present. C2F4 LaTeX: \longrightarrowⶠ1/2 C4F8 What is the half-life (in seconds) of the reaction for the initial C2F4 concentration? Enter to 1 decimal place.
Answer:
134.8 seconds is the half-life (in seconds) of the reaction for the initial [tex]C_2F_4[/tex] concentration
Explanation:
Half life for second order kinetics is given by:
[tex]t_{\frac{1}{2}=\frac{1}{k\times a_0}[/tex]
Integrated rate law for second order kinetics is given by:
[tex]\frac{1}{a}=kt+\frac{1}{a_0}[/tex]
[tex]t_{\frac{1}{2}[/tex] = half life
k = rate constant
[tex]a_0[/tex] = initial concentration
a = Final concentration of reactant after time t
We have :
[tex]C_2F_4 \longrightarrow \frac{1}{2} C_4F_8[/tex]
Initial concentration of [tex]C_2F_4=[a_o]=\frac{0.438 mol}{2.42 L}=0.1810 mol/L[/tex]
Rate constant = k = [tex]0.041 M^{-1} s^{-1}[/tex]
[tex]t_{\frac{1}{2}=\frac{1}{k\times a_0}[/tex]
[tex]=\frac{1}{0.041 M^{-1} s^{-1}\times 0.1810 mol/L}[/tex]
[tex]t_{1/2}=134.8 s[/tex]
134.8 seconds is the half-life (in seconds) of the reaction for the initial [tex]C_2F_4[/tex] concentration
The second-order rate constant for the following gas-phase reaction
C₂F₄ ⇒ 1/2 C₄F₈
is 0.0411 M⁻¹.s⁻¹. We start with 0.438 mol C₂F₄ in a 2.42-liter container, with no C₄F₈ initially present. What is the half-life (in seconds) of the reaction for the initial C₂F₄ concentration? Enter to 1 decimal place.
The reaction C₂F₄ ⇒ 1/2 C₄F₈, with a second-order rate constant of 0.0411 M⁻¹.s⁻¹, that starts with 0.438 mol C₂F₄ in a 2.42-liter container, has a half-life of 134.4 s.
Let's consider the following reaction following second-order kinetics.
C₂F₄ ⇒ 1/2 C₄F₈
Initially, we have 0.438 mol C₂F₄ in a 2.42-liter container. The initial concentration of C₂F₄ is:
[tex][C_2F_4]_0 = \frac{0.438 mol}{2.42 L} = 0.181 M[/tex]
Given the rate constant (k) is 0.041 M⁻¹.s⁻¹, we can calculate the half-life of a second-order reaction using the following expression.
[tex]t_{1/2}= \frac{1}{k \times [C_2F_4]_0} = \frac{1}{0.0411 M^{-1}s^{-1} \times 0.181 M} = 134.4 s[/tex]
The reaction C₂F₄ ⇒ 1/2 C₄F₈, with a second-order rate constant of 0.0411 M⁻¹.s⁻¹, that starts with 0.438 mol C₂F₄ in a 2.42-liter container, has a half-life of 134.4 s.
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