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ollegr [7]
3 years ago
8

A sample of an unknown gas with a mass of 8.21 g has a volume of 4.8064 L when the temperature is 200oC and the pressure is 1.81

6 atm. What is the molar mass of the gas?
Chemistry
1 answer:
Pie3 years ago
4 0

Answer:

The answer to your question is M = 36.49 g

Explanation:

Data

mass = 8.21 g

volume = 4.8064 L

Temperature = 200°C

Pressure = 1.816 atm

M = ?

Process

1.- Convert temperature to °K

°K = 273 + 200

°K = 473

2.- Calculate the number of moles

n = (PV)/RT

n = (1.816)(4.8064)/(0.082)(473)

n = 0.225

3.- Calculate the molar mass

            M --------------- 1 mol

           8.21 g ---------- 0.225 moles

M = (1 x 8.21)/0.225

M = 36.49 g

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Determine the pHpH of an HFHF solution of each of the following concentrations. In which cases can you not make the simplifying
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The question is incomplete, complete question is :

Determine the pH of an HF solution of each of the following concentrations. In which cases can you not make the simplifying assumption that x is small? (K_a for HF is 6.8\times 10^{-4}.)

[HF] = 0.280 M

Express your answer to two decimal places.

Answer:

The pH of an 0.280 M HF solution is 1.87.

Explanation:3

Initial concentration if HF = c = 0.280 M

Dissociation constant of the HF = K_a=6.8\times 10^{-4}

HF\rightleftharpoons H^++F^-

Initially

c          0            0

At equilibrium :

(c-x)      x             x

The expression of disassociation constant is given as:

K_a=\frac{[H^+][F^-]}{[HF]}

K_a=\frac{x\times x}{(c-x)}

6.8\times 10^{-4}=\frac{x^2}{(0.280 M-x)}

Solving for x, we get:

x = 0.01346 M

So, the concentration of hydrogen ion at equilibrium is :

[H^+]=x=0.01346 M

The pH of the solution is ;

pH=-\log[H^+]=-\log[0.01346 M]=1.87

The pH of an 0.280 M HF solution is 1.87.

6 0
3 years ago
A bond created from the sharing of electrons between two atoms is a(an) ______ bond.
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When 551. mg of a certain molecular compound X are dissolved in 100 g of benzonitrile (CH,CN), the freezing point of the solutio
arsen [322]

Answer:

1.12g/mol

Explanation:

The freezing point depression of a solvent for the addition of a solute follows the equation:

ΔT = Kf*m*i

<em>Where ΔT is change in temperature (Benzonitrile freezing point: -12.82°C; Freezing point solution: 13.4°C)</em>

<em>ΔT = 13.4°C - (-12.82) = 26.22°C</em>

<em>m is molality of the solution</em>

<em>Kf is freezing point depression constant of benzonitrile (5.35°Ckgmol⁻¹)</em>

<em>And i is Van't Hoff factor (1 for all solutes in benzonitrile)</em>

Replacing:

26.22°C = 5.35°Ckgmol⁻¹*m*1

4.90mol/kg = molality of the compound X

As the mass of the solvent is 100g = 0.100kg:

4.9mol/kg * 0.100kg = 0.490moles

There are 0.490 moles of X in 551mg = 0.551g, the molar mass (Ratio of grams and moles) is:

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2 years ago
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Answer:

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4 0
3 years ago
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2

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3 0
3 years ago
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