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stellarik [79]
3 years ago
6

The pressure exerted by 1.5 mol of gas in a 13 L flask at 22 °C is ____ kPa

Chemistry
1 answer:
sukhopar [10]3 years ago
8 0

Answer:

282.7KPa

Explanation:

Step 1:

Data obtained from the question.

Number of mole of (n) = 1.5 mole

Volume (V) = 13L

Temperature (T) = 22°C = 22 + 273°C = 295K

Pressure (P) =..?

Gas constant (R) = 0.082atm.L/Kmol

Step 2:

Determination of the pressure exerted by the gas.

This can be obtained by using the ideal gas equation as follow:

PV = nRT

P = nRT /V

P = 1.5 x 0.082 x 295 / 13

P = 2.79atm.

Step 3:

Conversion of 2.79atm to KPa.

This is illustrated below:

1 atm = 101.325KPa

Therefore, 2.79atm = 2.79 x 101.325 = 282.7KPa

Therefore, the pressure exerted by the gas in KPa is 282.7KPa

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RUDIKE [14]

Answer:

The question is incomplete. The response options are as follows:

I. C-O or C=O

II. C-C

III. C-H

IV. O-H

The answer is: IV>III>I>II

Explanation:

V) O-H is a hydrogen bridge. The hydrogen bridge is characterized by being similar to dipole-dipole bonds.

III) C-H is an ionic bond. The ionic bond occurs when they fuse together due to electron transfer.

I) C=O is a covalent bond. The covalent bond happens when two atoms bond together to create a molecule, sharing its electrons that are in its most superficial layer,

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3 years ago
Consider a galvanic cell consisting of the following two redox couplesAlB+ 3e Al , Mga + 2e Mg, (ag. E® = 1.676 V ) E° = 2.356 V
malfutka [58]

Answer:

0.68 V

Explanation:

For anode;

3Mg(s) ---->3Mg^2+(aq) + 6e

For cathode;

2Al^3+(aq) + 6e -----> 2Al(s)

Overall balanced reaction equation;

3Mg(s) + 2Al^3+(aq) ----> 3Mg^2+(aq) + 2Al(s)

Since

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4 0
3 years ago
The complete combustion of propane (C3H8) in the presence of oxygen yields CO2 and H2O: C3H8 (g) + 5O2 (g) 3CO2 (g) + 4H2O (g) a
mixas84 [53]

Answer:

26.9 L is the volume of CO₂, we obtained

Explanation:

The reaction is: C₃H₈(g) + 5O₂(g)  →  3CO₂ (g) + 4H₂O (g)

Let's determine the reactants moles:

27.5 g . 1mol / 44 g = 0.625 moles

We need density of O₂ to determine mass and then, the moles.

O₂ density = O₂ mass / O₂ volume

O₂ density . O₂ volume = O₂ mass

1.429 g/L . 45L = O₂ mass → 64.3 g

Moles of O₂ → 64.3 g . 1mol/32g = 2.009 moles

Let's find out the limiting reactant:

1 mol of propane needs 5 moles of oxygen to react

Then, 0.625 moles will react with (0.625 . 5)/1 = 3.125 moles of O₂

Oxygen is the limiting reactant, we need 3.125 moles but we only have 2.009 moles

Ratio is 5:3. 5 moles of O₂ produce 3 moles of CO₂

Therefore, 2.009 moles of O₂ must produce (2.009 .3) /5 = 1.21 moles of CO₂. Let's find out the volume, by Ideal Gases Law (STP are 1 atm and 273K, the standard conditions)

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3 0
3 years ago
PLZZZZZ HELP MEEEEEE
Masja [62]

Answer:

0.0745 mole of hydrogen gas

Explanation:

Given parameters:

Number of H₂SO₄ = 0.0745 moles

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Unknown:

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Solution:

To solve this problem, we have to work from the known specie to the unknown one.

The known specie in this expression is the sulfuric acid,  H₂SO₄. We can compare its number of moles with that of the unknown using a balanced chemical equation.

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   since 1 mole of sulfuric acid produced 1 mole of hydrogen gas;

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