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GrogVix [38]
3 years ago
9

The gases listed are at standard temperature and pressure: · 1 mol of CO2 gas · 1 mol of N2 gas · 2 mol of O2 gas Which statemen

t is correct? The volume of 1 mol of CO2 is greater than that of 2 mol of O2 . The volume of 1 mol of CO2 is greater than that of 1 mol of N2 . The volumes of 2 mol of O2 and 1 mol of N2 gases are the same. **The volumes of 1 mol of CO2 and 1 mol of N2 gases are the same.**
Chemistry
2 answers:
andrezito [222]3 years ago
8 0
The right answer for the question that is being asked and shown above is that: "The volumes of 1 mol of CO2 and 1 mol of N2 gases are the same." The <span>statement that is correct as far as the </span>gases listed are at standard temperature and pressure is concerned is that t<span>he volumes of 1 mol of CO2 and 1 mol of N2 gases are the same</span>
Elena L [17]3 years ago
6 0

Answer : The volume of 1 mole of CO_{2} and N_{2} gases are the same.

Solution: Given,

Moles of CO_{2} = 1 mole

Moles of N_{2} = 1 mole

Moles of O_{2} = 2 mole

Formula used for ideal gas is :

P V = n R T

According to the question, the gases are at standard temperature and pressure. So, the volume of gases only depends on the number of moles. This means that the higher the number of moles, higher will be the volume of gas.

The moles of O_{2} are more than the moles of CO_{2} and N_{2}. So, the volume of O_{2} will be more.

And the moles of CO_{2} and N_{2} are equal. Therefore, their volumes are also equal.

Therefore, the best option is the volume of 1 mole of CO_{2} and N_{2} gases are the same.

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A 0.1873 g sample of a pure, solid acid, H2X (a diprotic acid) was dissolved in water and titrated with 0.1052 M NaOH solution.
puteri [66]

Answer:

We need 41.8 mL of NaOH

Explanation:

<u>Step 1:</u> Data given

Mass of H2X = 0.1873 grams

Molarity of NaOH solution = 0.1052 M

Molar mass of H2X = 85.00 g/mol

<u>Step 2</u>: The balanced equation

H2X (aq) +2 NaOH (aq) → Na2X (aq) + 2H2O(l)

<u>Step 3:</u> Calculate moles of H2X

Moles H2X = mass H2X / Molar mass H2X

Moles H2X = 0.1873 grams / 85.00 g/mol

Moles H2X = 0.0022 moles

<u>Step 4:</u> Calculate moles of NaOH

For 1 mol H2X we need 2 moles NaOH to produce 1 mole of Na2X and 2 moles of H2O

For 0.0022 moles of H2X we need 0.0044 moles of NaOH

<u />

<u>Step 5</u>: Calculate volume of NaOH

Volume of NaOH = moles of NaOH / molarity of NaOH

Volume of NaOH = 0.0044 moles / 0.1052 M

Volume NaOH =  0.0418 L = 41.8 mL

We need 41.8 mL of NaOH

6 0
3 years ago
Find the percent ionization of a 0.337 m hf solution. the ka for hf is 3.5 x 10-4. 1.1 % 1.2 x 10-2 % 3.2 % 3.5 x 10-2 % 4.7 %
Ugo [173]

To determine the percent ionization of the acid given, we make use of the acid equilibrium constant (Ka) given. It is the ration of the equilibrium concentrations of the dissociated ions and the acid. The dissociation reaction of the HF acid would be as follows:<span>

HF = H+ + F-

The acid equilibrum constant would be expressed as follows:

Ka = [H+][F-] / [HF] = 3.5 x 10-4

To determine the equilibrium concentrations we use the ICE table,
         HF             H+              F-
I      0.337           0                 0
C      -x              +x               +x
---------------------------------------------
E    0.337-x        x                   x 

3.5 x 10-4 = [H+][F-] / [HF] 
3.5 x 10-4 = [x][x] / [0.337-x] </span>

Solving for x,

x = 0.01069 = [H+] = [F-]

percent ionization = 0.01069 / 0.337 x 100  = 3.17%

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