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Vsevolod [243]
2 years ago
8

A sample of CO2 gas at 100 degrees Celsius has a volume of 250 mL at 760 mm Hg. How many moles of CO2 are present

Chemistry
1 answer:
Fed [463]2 years ago
6 0

There are 8.16 × 10-³ moles of CO2 gas at 100°C with a volume of 250 mL at 760 mm Hg.

HOW TO CALCULATE NUMBER OF MOLES:

The number of moles of a sample of gas can be calculated using the following formula:

PV = nRT

Where;

  • P = pressure of gas (atm)
  • V = volume (L)
  • n = number of moles (mol)
  • R = gas law constant (0.0821 Latm/molK)
  • T = temperature (K)

According to this question;

  • P = 760mmHg = 1 atm
  • T = 100°C = 100 + 273 = 373K
  • V = 250mL = 0.250L
  • n = ?

1 × 0.250 = n × 0.0821 × 373

0.250 = 30.62n

n = 0.250 ÷ 30.62

n = 8.16 × 10-³mol

Therefore, there are 8.16 × 10-³ moles of CO2 gas at 100°C with a volume of 250 mL at 760 mm Hg.

Learn more about number of moles at: brainly.com/question/4147359

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For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)
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Answer : The value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

Explanation :

As we know that,

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = -45.6 kJ = -45600 J

\Delta S^o = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

Now put all the given values in the above formula, we get:

\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)

\Delta G^o=-12666.6J=-12.7kJ

The relation between the equilibrium constant and standard Gibbs free energy is:

\Delta G^o=-RT\times \ln k

where,

\Delta G^o = standard Gibbs free energy  = -12666.6 J

R = gas constant  = 8.314 J/K.mol

T = temperature  = 262.0 K

K = equilibrium constant = ?

Now put all the given values in the above formula, we get:

-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k

k=3.35\times 10^{2}

Therefore, the value of equilibrium constant for this reaction at 262.0 K is 3.35\times 10^{2}

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A fat is composed of long chains of carbon and hydrogen atoms. In a reaction with a strong base, a fat forms a soap and glycerol
Nonamiya [84]

Answer:

Empirical formula is C₉H₁₅O

Molecular formula = C₈₁H₁₃₅O ₉

Explanation:

Percentage of carbon = 77.60%

Percentage of oxygen = 11.45%

Percentage of hydrogen = 10.95%

Molecular weight = 1253 g/mol

Molecular formula = ?

Empirical formula = ?

Solution:

Number of gram atoms of C = 77.60 g /12g/mol =6.5

Number of gram atoms of O = 11.45 g / 16 g/mol = 0.72

Number of gram atoms of H = 10.95 g / 1.008 g/mol= 10.9

Atomic ratio:

C               :            H                 :    O

6.5/0.72   :       10.9/0.72         :   0.72/0.72

     9          :            15              :         1

C : H : O = 9 : 15 : 1

Empirical formula is C₉H₁₅O

Molecular formula:

Molecular formula = n (empirical formula)

n = molar mass of compound / empirical formula mass

n = 1253 / 139

n = 9

Molecular formula = n (empirical formula)

Molecular formula = 9 (C₉H₁₅O )

Molecular formula = C₈₁H₁₃₅O ₉

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