How many molecules of glucose (C6H12O6) are present in 23.3 g of the substance?

Answer:

B - End to end overlap of p overlap

Explanation:

Sigma bonds (σ bonds) are the strongest type of covalent chemical bondSigma bonds are formed by end-to-end overlapping. Sigma bonds can occur between any kind of atomic orbitals;The combination of overlapping orbitals can be s-s, s-pz or pz-pz.                              

Answer:

in the presence of water H2O

Na2CO3 (S) --> 2Na+ (aq)+ (CO3)2-(aq)

One mole of sodium carbonate produces two moles of Na+ ions

Therefore 0.207 moles produces 0.414 moles of Na+ ions

= 0.414 moles of Na+ ions

Explanation:

In water

Na2CO3 --> 2Na+ (aq)+ (CO3)2-(aq)

In a limited reaction, the carbonate ion reacts with the water molecules as follows

(CO3)2-(aq) + H2O←→HCO3-(aq) + OH-(aq)

sodium carbonate or soda ash dissolves in water to give 2 sodium cations and one carbonate anion

The balanced equation for the dissolution of sodium carbonate in water produces two moles of Na+ for every mole of Na2CO3. For 0.207 mol of sodium carbonate, it yields 0.414 mol of Na+ ions.

The balanced equation for the dissolution of sodium carbonate (Na2CO3) in water is:

Na2CO3(s) → 2Na+(aq) + CO3^2-(aq)

When 0.207 mol of sodium carbonate dissolves in water, it produces two moles of Na+ ions for every mole of sodium carbonate. Therefore, to find the number of moles of Na+ produced we multiply:

0.207 mol Na2CO3 × 2 mol Na+ / 1 mol Na2CO3 = 0.414 mol Na+

Answer:

moles B = 2.32 moles

Explanation:

In this case, we can assume that both gases are ideals, so we can use the expression for an ideal gas which is:

PV = nRT

From here, we can calculate the total moles (n) that are in the container, and then, by difference, we can calculate how much we have of gas B.

For this case, we will use R = 0.082 L atm / mol K. Solving for n:

n = PV/RT

n = 5 * 20 / 0.082 * 303

n = 4.02 moles

If we have 4.02 moles between the two gases, and we have 1.70 from gas A, then from gas B we simply have:

Total moles = moles A + moles B

moles B = Total moles - moles A

moles B = 4.02 - 1.70

moles B = 2.32 moles

We have 2.32 moles of gas B

Considering the ideal gas law, 2.32 moles of Gas B are present in the mixture.

Ideal gases are a simplification of real gases that is done to study them more easily. It is considered to be formed by point particles, do not interact with each other and move randomly. It is also considered that the molecules of an ideal gas, in themselves, do not occupy any volume.

The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:  

P×V = n×R×T

where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.

In this case, you know:

Replacing in the ideal gas law:

5 atm× 20 L= n× 0.082[tex]\frac{atmL}{molK}[/tex]× 303 K

Solving:

[tex]n=\frac{5 atmx20 L}{0.082\frac{atmL}{molK}x303 K }[/tex]

n= 4.02 moles

You have 4.02 moles between the two gases, and you have 1.70 from gas A. Then the number of moles of gas B can be calculated as:

Total moles = moles A + moles B

4.02 moles= 1.70 moles + moles B

4.02 moles - 1.70 moles= moles B

2.32 moles= moles B

Finally, 2.32 moles of Gas B are present in the mixture.

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Answer:

a. the negative ends of water molecules surround the positive ions.

Explanation:

Ionization reaction is defined as the reaction in which an ionic compound dissociates into its ions when dissolved in aqueous solution.

Thus,

[tex]NaCl_{(s)}\rightarrow Na^+_{(aq)}+Cl^-_{(aq)}[/tex]

When the ionic compound dissociates, the cation which is positively charged gets attracted to the oxygen (Negative ends) which is present in the water. The anion which is negatively charged gets attracted to the hydrogen (Positive ends) of the water.

Hence, the correct option is:- a. the negative ends of water molecules surround the positive ions.

When an ionic compound dissolves in water, the positive end of the water molecule surrounds the negative ions, and the negative end of the water molecule surrounds the positive ions.

When an ionic compound dissolves in water, the water molecules surround the individual ions and pull them apart through a process known as dissolving or hydration. Specifically, the positive end of the water molecule (which is the hydrogen end) surrounds the negative ions, while the negative end of the water molecule (which is the oxygen end) surrounds the positive ions. So, the correct answer is: (a) the negative ends of the water molecules surround the positive ions and (c) the positive ends of water molecules surround the negative ions. Therefore, water is described as a polar solvent, due to its capability to dissolve many ionic substances.

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Answer:

[HCl] = 2.77 M   (option b)

Explanation:

A 10% by mass is an information that means, 10 g of solute are contained in 100 g of solution.

Density is mass / volume and it always corresponds to solution.

So solution mass is 100 g. Let's find out solution volume with density

Solution mass / Solution volume = Solution density

100 g / Solution volume = 1.01 g/mL

100 g / 1.01 g/mL = 99 mL

In conclusion our 10 g of HCl are contained in 99 mL of solution

Molarity is mol of solute in 1L of solution

Let's convert the mass to moles (mass / molar mass)

10 g / 36.45 g/mol = 0.274 moles

Let's convert the mL in L

99 mL = 0.099L

Molarity is mol/L → 0.274 mol / 0.099L = 2.77M

Final answer:

To calculate the molarity of a 10.0% (by mass) aqueous solution of hydrochloric acid, you can follow a step-by-step procedure. The molarity of the solution is approximately 2.77 M.

Explanation:

To calculate the molarity of a 10.0% (by mass) aqueous solution of hydrochloric acid, we need to first determine the amount of HCl in grams in the solution. Then, we convert the grams of HCl to moles using its molar mass. Finally, we divide the moles of HCl by the volume of the solution in liters to obtain the molarity.

Given that the solution has a density of 1.01 g/mL and is 10.0% HCl by mass, we can assume a 100 g sample of the solution, which would contain 10 g of HCl. Converting the volume of the solution to liters, we have 100 g ÷ 1.01 g/mL = 99.01 mL ÷ 1000 mL/L = 0.09901 L.

Now, we calculate the moles of HCl: moles = grams ÷ molar mass. The molar mass of HCl is approximately 36.46 g/mol. So, moles = 10 g ÷ 36.46 g/mol ≈ 0.274 mol. Finally, we divide moles by volume (in liters) to get the molarity: Molarity = 0.274 mol ÷ 0.09901 L ≈ 2.77 M.

Answer: The molecular formula of the compound is [tex]C_3H_{12}O_3[/tex]

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C = 37.5 g

Mass of H = 12.5 g

Mass of O = 50.0 g

Step 1 : convert given masses into moles.

Moles of C =[tex]\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{37.5g}{12g/mole}=3.125moles[/tex]

Moles of H = [tex]\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{12.5g}{1g/mole}=12.5moles[/tex]

Moles of O = [tex]\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{50.0g}{16g/mole}=3.125moles[/tex]

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C =[tex]\frac{3.125}{3.125}=1[/tex]

For H= [tex]\frac{12.5}{3.125}=4[/tex]

For O =[tex]\frac{3.125}{3.125}=1[/tex]

The ratio of C: H: O = 1: 4: 1

Hence the empirical formula is [tex]CH_4O[/tex].

The empirical weight of [tex]CH_4O[/tex] = 1(12)+4(1)+1(16)= 32g.

The molecular weight = 93.0 g/mole

Now we have to calculate the molecular formula.

[tex]n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{93}{32}=3[/tex]

The molecular formula will be=[tex]3\times CH_4O=C_3H_{12}O_3[/tex]

The molecular formula of the compound is C3H12O3. This was determined by converting percent composition to grams, then to moles to find a simplest ratio for the empirical formula and comparing its mass with the molecular mass.

The compound could firstly be analyzed into its empirical formula by treating the percentage composition directly as masses, which would then provide us the simplest ratio of elements. Hence for 37.5g of C we have 37.5/12.01 = ~3.1 moles of C, for 12.5g of H we have 12.5/1.01 = ~12.4 moles of H, and for 50.0g of O we have 50.0/16.00 = ~3.1 moles of O. The simplest ratio (empirical formula) will then be C3H12O3, with empirical formula mass equals to 94.1 g/mol. This mass is very close to the mentioned molecular mass of 93.0 g/mol, so our molecular formula is presumably the same as the empirical formula. Therefore, the molecular formula of the compound is C3H12O3.

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Answer:

Drift speed of electrons will be 1.056x10^-4 m/s

Given Data:

A(area)= 5.4 x 10^-6 [tex]m^{2}[/tex]

I(current)= 5.5 A

Density= 2.7 [tex]g/cm^{3}[/tex]

The equation for drift velocity is:

[tex]v(drift)=I/nqA[/tex]

In this case 'q' will be charge of electron which is= 1.6 x 10-19

As each atoms supplies one conduction electron, so number of conduction electrons will be equal to number of atoms.

Hence,

n= no. of conduction electrons/[tex]m^{3}[/tex] = no. of atoms/[tex]m^{3}[/tex]

To find 'n' we can use following equation:

[tex]n= (mass/cm^{3} *atoms/mol)/(mass/mol)[/tex]

We know atoms/mol is equal to Avogadro`s number i.e 6.02 x 10^23

and molar mass of aluminium is 26.982 g.

Now,

[tex]n=(2.7g/cm^{3} * 6.02*10^{23} )/25.982g[/tex]    (putting values in above equation)

[tex]n=6.024*10^{22} electrons/cm^{3}[/tex]

[tex]n= 6.024*10^{22} *10^{6} electrons/m^{3}[/tex] (converting electrons/cm3 to electrons/m3)

[tex]n= 6.024*10^{28} electrons/m^{3}[/tex]

To find drift velocity, we will use equations mention before:

[tex]v(drift)=I/nqA[/tex]

[tex]v(drift)=5.5A/(6.024*10^{28}electrons/m^{3} *1.6*10^{-19}C* 5.4*10^{-6}m^{2} )[/tex]

[tex]v(drift)= 1.056*10^{-4} m/s[/tex]

The drift velocity of electrons in the aluminum wire is 1.353 × 10^-3 m/s.

The drift velocity of electrons in a wire can be calculated using the formula Vd = I / (nqA), where Vd is the drift velocity, I is the current, n is the number of free electrons per unit volume, q is the charge of an electron, and A is the cross-sectional area of the wire.

In this case, the cross-sectional area of the aluminum wire is given as 5.40 × 10-6 m2 and the current is 5.50 A. The charge of an electron is -1.60 × 10-19 C. To calculate the number of free electrons per unit volume, we need to know the density of aluminum and the molar mass of aluminum.

Using the given data, the drift velocity of the electrons in the aluminum wire can be calculated as follows:

Vd = (5.50 A) / [(8.44 × 1028 m-3) × (-1.60 × 10-19 C) × (5.40 × 10-6 m2)] = 1.353 × 10-3 m/s.

Answer:

The answer to your question is Yes, it is a double replacement reaction

Explanation:

General equation of a double replacement reaction

                         AB  +  CD   ⇒   AD  +   BC

In a double replacement reaction, the cation of a compound is interchanged with the cation and another compound and also the anions are interchanged. The reaction given has the characteristics of a double replacement reaction.

Answer:

A. Covalent bond

B. Hydrogen bond

Explanation:

Covalent bonds form within a water molecule between the oxygen and hydrogen atoms, while hydrogen bonds form between different water molecules due to electrical attraction.

In a water molecule, covalent bonds form between the oxygen and hydrogen atoms. This occurs because they share electrons to fulfill the Octet Rule for stable electron configurations. On the other hand, when different water molecules interact with each other, a form of weak connection known as hydrogen bonds are formed. This happens as the negatively charged oxygen end of one water molecule is attracted to the positively charged hydrogen end of another water molecule. This type of bonding results in some unique properties of water, such as its ability to act as a solvent for many substances.

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Answer: The electronic configuration of the elements are written below.

Explanation:

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.

For the given options:

Carbon is the 6th element of the periodic table. The number of electrons in carbon atom are 6.

The electronic configuration of carbon is [tex]1s^22s^22p^2[/tex]

Phosphorus is the 15th element of the periodic table. The number of electrons in phosphorus atom are 15.

The electronic configuration of phosphorus is [tex]1s^22s^22p^63s^23p^3[/tex]

Vanadium is the 23rd element of the periodic table. The number of electrons in vanadium atom are 23.

The electronic configuration of vanadium is [tex]1s^22s^22p^63s^23p^64s^23d^3[/tex]

Antimony is the 51st element of the periodic table. The number of electrons in antimony atom are 51.

The electronic configuration of antimony is [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^3[/tex]

Samarium is the 62nd element of the periodic table. The number of electrons in samarium atom are 62.

The electronic configuration of samarium is [tex]1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^24f^6[/tex]

Hence, the electronic configuration of the elements are written above.

The electrons configurations of the elements C, P, V, Sb, and Sm were determined using complete subshell notation. It depicts the filling of electrons into s, p, d, and f orbitals at different energy levels according to the Aufbau Principle.

The electron configurations in complete subshell notation for the elements specified (C, P, V, Sb, and Sm) would be as following:

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The mass of oxygen reacted with 14 g nitrogen has been 16 grams.

The formation of dinitrogen dioxide has been mediated with the reaction of oxygen with nitrogen.

The mass of oxygen required for the reaction has been given with the balanced chemical equation as:

[tex]\rm N_2\;+\;O_2\;\rightarrow\;N_2O_2[/tex]

Thus, for the reaction with 1 mole of oxygen, 1 mole of nitrogen has been required.

The mass of 1 mole oxygen has been 32 g.

The mass of 1 mole nitrogen has been 28 g.

Thus, for the reaction with 14 g nitrogen, the mass of oxygen required has been:

[tex]\rm 28\g\;N_2=32\;g\;O_2\\14\;g\;N_2=\dfrac{32}{28}\;\times\;14\g\;O_2\\14\;g\;N_2=165\;g\;O_2[/tex]

The mass of oxygen reacted with 14 g nitrogen has been 16 grams.

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The mass of oxygen required to react with 14 g of nitrogen to make [tex]N_2O_5[/tex] is 80 g.

The balanced chemical equation for the formation of [tex]N_2O_5[/tex] from nitrogen [tex](N_2)[/tex] and oxygen [tex](O_2)[/tex] is:

[tex]\[ N_2 + 5O_2 \rightarrow 2N_2O_5 \][/tex]

The balanced chemical equation for the formation of [tex]N_2O_2[/tex] from nitrogen [tex](N_2)[/tex] and oxygen [tex](O_2)[/tex] is:

[tex]\[ N_2 + 2O_2 \rightarrow 2N_2O_2 \][/tex]

From the second equation, we can see that 16 g of oxygen reacts with 14 g of nitrogen to form [tex]N_2O_2[/tex]. This gives us the molar ratio of oxygen to nitrogen for [tex]N_2O_2[/tex], which is 2:1 by mass.

Now, we need to find out the molar ratio of oxygen to nitrogen for [tex]N_2O_5[/tex]. From the first equation, the molar ratio of oxygen to nitrogen is 5:1.

Since the mass of nitrogen (14 g) is the same in both reactions, we can directly compare the ratios of oxygen used in both reactions.

For [tex]N_2O_2[/tex], we have 16 g of oxygen for every 14 g of nitrogen, and for [tex]N_2O_5[/tex], we need 5 times as much oxygen for the same amount of nitrogen.

Therefore, the mass of oxygen required for [tex]N_2O_5[/tex] is:

[tex]\[ \frac{5}{2} \times 16 \text{ g of oxygen} = 40 \text{ g of oxygen} \][/tex]

Using the molar masses of nitrogen (14 g/mol for [tex]N_2[/tex]) and oxygen (32 g/mol for [tex]O_2[/tex]), we can calculate the amount of oxygen needed:

[tex]\[ \frac{14 \text{ g of N2}}{28 \text{ g/mol of N2}} \times \frac{5 \text{ mol of O2}}{1 \text{ mol of N2}} \times 32 \text{ g/mol of O2} = 80 \text{ g of O2} \][/tex]

Answer:

The answer to your question is   C₃H₃O

Explanation:

Data

Combustion of a compound C, H, O

mass = 6.10 g

mass CO2 = 14.9 g

mass of water = 6.10 g

Reaction

                     Cx Hy Oz   +   O2   ⇒     CO2   +   H2O

Process

1.- Calculate the moles of carbon

                          44 g of CO2   --------------  12 g of carbon

                           14.9 g of CO2 -------------   x

                            x = (14.9 x 12) / 44

                            x = 4.06 g

                          12 g of C    ------------------ 1 mol

                          4.06 g       ------------------- x

                          x = (4.06 x 1) / 12

                          x = 0.34 moles

2.- Calculate the moles of hydrogen

                           18 g of water -------------  1 g of hydrogen

                            6.10 g of water ----------   x

                            x = (6.10 x 1) / 18

                            x = 0.33 g

                           1 g of H  ----------------  1 mol of H

                           0.33 g     ----------------  x

                           x = (0.33 x 1) / 1

                           x = 0.33 moles of H

3.- Calculate the mass of oxygen

mass of Oxygen = 6.10 - 4.06 - 0.33

                            = 1.71 g

                          16 g of O ---------------  1 mol of O

                          1.71 g of O -------------   x

                          x = (1.71 x 1) / 16

                          x = 0.11 moles

4.- Divide by the lowest number of moles

Carbon   = 0.34 / 0.11  = 3                          

Hydrogen = 0.33 / 0.11 = 3

Oxygen = 0.11 /0.11 = 1

5.- Write the empirical formula

                                    C₃H₃O

The ship meant to transport the 'terranauts' to the Earth's core could potentially be made of a hypothetical material, like 'Unobtainium', capable of enduring extreme conditions.

The material of the ship transporting the terranauts to the core is not explicitly mentioned, but the answer would depend on the context provided by the source material (for instance, a science-fiction story or movie). In general, however, a material capable of withstanding extreme temperatures and pressures would be required for such a journey. This could potentially be a hypothetical, highly resilient element, such as Unobtainium (D).

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Answer: 0.13mol

Explanation:Please see attachment for explanation

Answer:aluminum develops a coating of aluminum oxide,

Explanation:

Aluminium has an electrode potential value of -1.66V while iron has an electrode potential value of -0.44V for iron II and +0.77 V for iron III. Clearly, aluminum has a more negative value of electrode potential and ought to be more reactive. However, a protective coating formed on aluminium surface prevents corrosion of the metal.

Answer : The value correspond to in miles per gallon is, 20.6976 mile/gallon

Explanation :

The conversion used to convert kilometer to miles is:

1 km = 0.6214 miles

The conversion used to convert liter to gallon is:

1 L = 0.2642 gallons

Thus,

1 km/L = [tex]\frac{0.6214}{0.2642}mile/gallon[/tex]

1 km/L = 2.352 mile/gallon

As we are given that 8.8 km/L of gasoline. Now we have to convert it into mile/gallon.

As, 1 km/L = 2.352 mile/gallon

So, 8.8 km/L = [tex]\frac{8.8km/L}{1km/L}\times 2.352mile/gallon[/tex]

                     = 20.6976 mile/gallon

Thus, the value correspond to in miles per gallon is, 20.6976 mile/gallon

To convert the fuel economy from km/L to mpg, multiply the km/L value by 2.8248.

To convert the fuel economy from kilometers per liter (km/L) to miles per gallon (mpg), we can use the conversion factor:

1 km/L = 2.8248 mpg

So, to find the fuel economy in miles per gallon, we multiply 8.8 km/L by 2.8248:

8.8 km/L × 2.8248 mpg = 24.89904 mpg

Therefore, the fuel economy of the rental car is approximately 24.9 miles per gallon.

Answer:

0.15 m C6H12O6 > 0.15 m CH3COOH > 0.10 m Na3PO4 >  0.20 m MgCl2 > 0.35 m NaCl

Explanation:

Step 1:

0.10m Na3PO4

Na3PO4 → 3Na+ + PO4^-3-

⇒ van't Hoff factor is 3 + 1 = 4

0.10 m * 4 = 0.40

0.35m NaCl

NaCl → Na+ + Cl-

⇒ van't Hoff factor = 1+1 = 2

0.35 * 2 = 0.70

0.20m MgCl2

MgCl2 → Mg^2+ + 2Cl-

 ⇒ Van't Hoff factor = 1+2 = 3

0.20 * 3 = 0.60

0.15m C6H12O6

⇒  for non-ionic compounds in solution, like glucose (C6H12O6) , the van't Hoff factor is 1. They do not dissociate in water.

0.15 * 1 = 0.15

0.15m CH3COOH.

CH3CO2H ⇄ CH3CO2− + H+

 ⇒ Van't hoff factor ≈ 1<x<2

0.15 * 2 = 0.30

Larger the concentration of the concentration of the particles, smaller the freezing point.

0.15 m C6H12O6 > 0.15 m CH3COOH > 0.10 m Na3PO4 >  0.20 m MgCl2 > 0.35 m NaCl

The order of decreasing freezing point is -

0.15m C6H12O6 < 0.15m CH3COOH < 0.10 m Na3PO4 < 0.20 m MgCl2 < 0.35 m NaCl,

It is calculated as:  

dT = i Kf m  

m = The concentration values in molalities  

Kf = the cryoscopic constant for water (-1.86, the same for all solutions)

i = the Van’t Hoff factor, which is the number of ion particles per individual molecule/formula of solute

Solution:

So you have to arrange in increasing order of the product i·m  

0.15 m C6H12O6        

im = 1 x 0.15 = 0.15     (undissociated, i=1)

0.15 m CH3COOH  

im =1.1 x 0.15 = 0.17 (partially dissociated) ( 1<i<2)

0.10 m Na3PO4

im = 4 x 0.1 = 0.4       (i=4)

0.20 m MgCl2

im = 3 x 0.2 = 0.6    (i=3)

0.35 m NaCl,

im = 2 x 0.35 = 0.7 (i=2)

Thus, the order of decreasing freezing point is -

0.15m C6H12O6 < 0.15m CH3COOH < 0.10 m Na3PO4 < 0.20 m MgCl2 < 0.35 m NaCl,

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Answer:

The answer to your question is 8.28 g of glucose

Explanation:

Data

Glucose  (C₆H₁₂O₆) = ?

Ethanol (CH₃CH₂OH)

Carbon dioxide (CO₂) = 2.25 l

Pressure = 1 atm

T = 295°K

Reaction

                             C₆H₁₂O₆    ⇒    2C₂H₅OH(l) +2CO₂(g)

- Calculate the number of moles

                             PV = nRT

Solve for n

                             [tex]n = \frac{PV}{RT}[/tex]

Substitution

                             [tex]n = \frac{(1)(2.25)}{(0.082)(295)}[/tex]

Simplification

                            n = 0.092

- Calculate the mass of glucose

                         1 mol of glucose --------------- 2 moles of carbon dioxide

                          x                         --------------- 0.092 moles

                          x = (0.092 x 1) / 2

                          x = 0.046 moles of glucose

Molecular weight of glucose = 180 g

                       180 g of glucose ---------------  1 mol

                         x  g                    ---------------0.046 moles

                         x = (0.046 x 180) / 1

                         x = 8.28 g of glucose                                    

Answer:

131.2 mL would be the total volume in the graduated cylinder.

Explanation:

Let's work with the density of metal to find out its volume

Metal density = Metal mass / metal volume

4.46 g/mL = 22.25 g / metal volume

Metal volume = 22.25 g / 4.46 g/mL = 4.98 mL

If we suppose aditive volume, total volume in the graduated cylinder will be:

126.3mL + 4.9mL =  131.2 mL

Answer:

[(172 - 170)mg/170 mg]*100% = 2*100%/170 = 1.18%

Explanation:

Generally percent error is calculated by dividing the error by the actual value of the variable at standard conditions. In the problem above, the acceptable or actual mass of potassium in a cup of the cereal is 170 mg and the estimated mass of potassium in a cup of the cereal is 172 mg. Therefore, the percent error = [(172 - 170)mg/170 mg]*100% = 2*100%/170 = 1.18%

In the molecule ammonia (NH3), each hydrogen atom has a partial positive charge and the nitrogen atom has a partial negative charge. There are covalent bonds between the hydrogen atoms.

In the molecule ammonia (NH3), each hydrogen atom has a partial positive charge (A) because nitrogen is more electronegative than hydrogen. The nitrogen atom, on the other hand, has a partial negative charge (D) due to the uneven distribution of electrons. The presence of covalent bonds between nitrogen and hydrogen atoms (E) results in the formation of ammonia.

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Answer:

5.38 moles of CO₂ are produced

Explanation:

This is the reaction:

C₃H₈  +  5O₂  →  3CO₂  +  4H₂O

First of all, let's convert the mass of C₃H₈ to moles (mass / molar mass)

79 g / 44 g/mol = 1.79 moles

So ratio is 1:3.

1 mol of C₃H₈ is needed to produce  3 moles of CO₂

1.79 moles of C₃H₈ would produce (1.79  .3) /1 =  5.38 moles

Explanation:

The Lewis dot diagram shows how electrons participate in a bond with Carbon and Chlorine. This is shown by the sticks and the 2 paired electrons near the carbon atom which represent the bonds. These electrons form these bonds because they form octets when they are bonded which most molecules and compounds follow

Hoped this helped, 2Trash4U

Answer: Chlorine and Carbon exhibits covalent bonding which involves the sharing if electrons. Carbon is less electronegative than Chlorine so it is considered as the central atom on this compound. Carbon shares it 4 valence electrons on Cl to attain octet. Now Chlorine is more electronegative so it will receive the lone pairs based from the Lewis structure.

Since C os in group 4 and Cl in group 7 the number of electrons will be:

C = 4

Cl= 7 x 4

# of e- = 32 - 8 ( the number of bonds)

Total number of e-. = 26 e-

Distribute the electrons to the most electronegative atom which is Cl have 6 e- each to attain octet.

Explanation:

Answer:

CHO2- ion

Explanation:

We have the lewis structure of a formate-ion here

This is CHO2-.

The carbon atom is the central atom in the structure. It's the least electronegative atom (C). Carbon has 4 valence electrons. Oxygen has 6 valence electrons.

The carbon will bind with 1 hydrogen atom, this will form 1 single bond, because hydrogen has 1 valence electron.

The carbon will bind with oxygen via a double bond.

Since carbon has only 4 valence electrons, it can only form 1 bond with the other oxygen atom.

There will formed 1 double bond between C and O and 1 single bond between C and O resulting in a negative charged O-atom.

This means there are two resonance structures. for the CHO2-  ion

Organic chemistry is a branch of chemistry studying carbon-containing compounds. While it traditionally focused on naturally occurring compounds, it now includes human-made substances. Applications range from pharmaceuticals to plastics.

Organic chemistry is a sub-discipline of chemistry that involves the study of the structure, properties, composition, reactions, and preparation of carbon-containing compounds, which includes not only hydrocarbons but also compounds with any number of other elements, including hydrogen (most compounds contain at least one carbon-hydrogen bond), nitrogen, oxygen, halogens, phosphorus, silicon, and sulfur.

This branch of chemistry was traditionally limited to compounds produced by living organisms but has been broadened to include human-made substances such as plastics. The range of application of organic compounds is enormous and also includes, but is not limited to, pharmaceuticals, dyes, coatings, plastics, and many more.

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Answer:

ΔHacetone = - 247.5 kJ/mol

Explanation:

The enthalpy equation is as follows

ΣnΔHproducts – ΣmΔHreactants =ΔHreaction

3×ΣnΔH(CO2(g)) + 3×ΣnΔH(H2O) - ΣnΔH (C3H6O(l)) =ΔHreaction = -1790 kJ/mol

[3(-393.5) + 3(-285.8)] – ΔHacetone

= -1790 kJ/mol

(-1180.5 – 857.4)kJ/mol - ΔHacetone =

-1790 kJ/mol

-2037.9 kJ/mol - ΔHacetone

= -1790kJ/mol

-2037.9 kJ/mol + 1790kJ/mol = ΔHacetone

- 247.5 kJ/mol = ΔHacetone

ΔHacetone = - 247.5 kJ/mol

To determine the standard enthalpy of formation of acetone (C3H6O), we can use the combustion reaction equation and the standard enthalpies of formation of other compounds involved. By applying Hess's Law and the enthalpy change of the reaction, we can calculate the standard enthalpy of formation.

The standard enthalpy of formation of acetone (C3H6O) can be calculated using the given equation for the combustion of acetone and the standard enthalpies of formation of O2(g), CO2(g), and H2O(l). For the combustion reaction:

C3H6O(l) + 4O2(g) → 3CO2(g) + 3H2O(l)

The enthalpy change or ΔH of the reaction is -1790 kJ. Using Hess's Law and the enthalpy change, we can determine the standard enthalpy of formation of acetone.

We can set up an equation using the standard enthalpy of formation values:

ΔH = Σ(ΔHf(products)) - Σ(ΔHf(reactants))

By rearranging the equation, we can solve for the standard enthalpy of formation of acetone.

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Answer:

6049.69 g or 6.04969 Kg of ore must be processed to get 1 Kg of phosphorous.

Explanation:

Given

An ore has 51% Calcium phosphate

To find

The minimum mass of ore to be processed to get 1.00 Kg of phosphorous

First find the mass of phosphorous in 1 mole = molar mass of Calcium phosphate, Ca₃(PO₄)₃.

Molar mass  of Ca₃(PO₄)₃ is 310 g

Molar mass  of  P is 31 g

1 mole of Ca₃(PO₄)₃ has 3 atoms of phosphorous

i.e 310 g of  Ca₃(PO₄)₃ has 3× 31g of P

= 93 g P

93 g P wil be present in 310 g of Ca₃(PO₄)₃  

1 Kg or 1000 g of P will be in   (1000÷93) ×310

=3333.33 g of Ca₃(PO₄)₃

but the ore has only 55.1% Ca₃(PO₄)₃

i.e

100 g of Ca₃(PO₄)₃ will have 55.1g Ca₃(PO₄)₃

we need 3333.33g of Ca₃(PO₄)₃

100 g of ore  will have 55.1g Ca₃(PO₄)₃

3333.33g of Ca₃(PO₄)₃ will be present in

(3333.33÷ 55.1) ×100

= 6049.60 g of the ore

So 6049.69 g or 6.04969 Kg of ore must be processed to get 1 Kg of phosphorous.

Answer:

6,216.684 kilograms of sodium carbonate must be added to neutralize [tex]6.05\times 10^3 kg[/tex] of sulfuric acid solution.

Explanation:

Mass of sulfuric acid solution = [tex]6.05\times 10^3 kg=6.05\times 10^6 g[/tex]

[tex]1 kg = 10^3 g[/tex]

Percentage mass of sulfuric acid = 95.0%

Mass of sulfuric acid = [tex]\frac{95.0}{100}\times 6.05\times 10^6 g[/tex]

[tex]=5,747,500 g[/tex]

Moles of sulfuric acid = [tex]\frac{5,747,500 g}{98 g/mol}=58,647.96 mol[/tex]

[tex]H_2SO_4+Na_2CO_3\rightarrow Na_2SO_4+CO_2+H_2O[/tex]

According to reaction , 1 mole of sulfuric acid is neutralized by 1 mole of sodium carbonate.

Then 58,647.96 moles of sulfuric acisd will be neutralized by :

[tex]\frac{1}{1}\times 58,647.96 mol=58,647.96 mol[/tex] of sodium carbonate

Mass of 58,647.96 moles of sodium carbonate :

[tex]106 g/mol\times 58,647.96 mol=6,216,683.76 g[/tex]

6,216,683.76 g = 6,216,683.76 × 0.001 kg = 6,216.684 kg

6,216.684 kilograms of sodium carbonate must be added to neutralize [tex]6.05\times 10^3 kg[/tex] of sulfuric acid solution.

To neutralize 6.05×10^3 kg of sulfuric acid solution, we would need approximately 6.21×10^3 kg of sodium carbonate, based on the stoichiometry of the chemical reaction between these two compounds.

The problem involves a chemical reaction between sulfuric acid (H2SO4) and sodium carbonate (Na2CO3). The balanced equation for this reaction is: H2SO4 + Na2CO3 -> Na2SO4 + H2O + CO2. From this equation, we can see that 1 mole of H2SO4 reacts with 1 mole of Na2CO3. Thus, the mass of Na2CO3 needed can be obtained by first determining the mass of H2SO4 present. Then, we convert the mass of H2SO4 to moles using its molar mass, and finally use the molar mass of Na2CO3 to find the mass of Na2CO3 needed.

First, we calculate the mass of H2SO4 in the solution: mass_H2SO4 = 0.95 * (6.05×10^3 kg of H2SO4 solution) = 5747.5 kg of H2SO4.

To convert this mass into moles, we divide by the molar mass of H2SO4, which is approximately 98.08 g/mol (or 0.09808 kg/mol). So, moles_H2SO4 = mass_H2SO4 / molar_mass_H2SO4 = 5747.5 kg / 0.09808 kg/mol = 58.6×10^3 mol.

From the balanced chemical equation, we know that 1 mole of H2SO4 neutralizes 1 mole of Na2CO3. Thus, to neutralize all the H2SO4, we need 58.6×10^3 mol of Na2CO3. Multiplying this number of moles by the molar mass of Na2CO3 (approximately 105.988 g/mol or 0.105988 kg/mol), we obtain the mass of Na2CO3 required: mass_Na2CO3 = moles_Na2CO3 * molar_mass_Na2CO3 = 58.6×10^3 mol * 0.105988 kg/mol = 6.21×10^3 kg of Na2CO3.

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Answer:

c.

Explanation:

A serial dilution is a dilution that is made fractionated. The stock solution is diluted, then this now solution is diluted, and then successively. The final dilution is the multiplication of the steps dilutions.

The representation of the dilution is v/v (volume per volume) indicates how much of the stock solution is in the total volume of the solution. So 1/5 indicates 1 mL to 5 mL of the solution. If the final volume must be 1 mL, then the stock solution must be 0.2 mL (0.2/1 = 1/5), and the volume of the solvent is 1 mL - 0.2 = 0.8 mL.

The second solution is done with a dilution of 1/10 or 1 mL of the first solution in 10 mL of the total solution. Because the solution has 1 mL, then the volume of the first solution must be 0.1 mL (0.1/1 = 1/10), and the volume of the solvent that must be added is 1 mL - 0.1 mL = 0.9 mL.

Final answer:

Other cation channels, such as voltage-gated calcium and potassium channels, can substitute for the function of blocked voltage-gated sodium channels in action potential generation and propagation.

Explanation:

If the voltage-gated sodium channels are blocked, other cation channels can substitute for their function in action potential generation and propagation. One such cation channel is the voltage-gated calcium channel. Although calcium channels primarily play a role in synaptic transmission, they can also contribute to the depolarization phase of the action potential in the absence of functional sodium channels.

Another cation channel that can substitute for voltage-gated sodium channels is the voltage-gated potassium channel. While potassium channels primarily repolarize the membrane during the action potential, they can also contribute to the depolarization phase in the absence of sodium channels.

It is important to note that these cation channels may not fully replicate the function of voltage-gated sodium channels, and the exact impact on action potential generation and propagation can vary depending on the specific circumstances.

Answer:

c) .51835

Explanation:

Let the relative abundance of the lighter of the two isotopes be X we have

Then the relative abundance of the heavier isotope is then (1-X)

Whereby we have that in nature the amount of the lighter silver found in proportion is X and the heavier isotope of silver is present as (1-X) proportion in nature.

To calculate the relative atomic mass of silver, we have

(Mass of light weight silver)×X + (mass of heavier isotope of silver×(1-X) = relative atomic mass of silver

106.90509(X) + 108.9047(1-X)

108.9-108.9(x)+106.9(x) = 107.87

-2x-1.03 = 0.517450902926

Closest answer is c

c) .5184

The relative atomic mass of isotopes is the weighted average by the mole-fraction of abundance of these isotopes which gives the atomic weight that is listed for that element on the periodic table.