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2 years ago

What is the mass in grams of 14.2 liters of sulfur dioxide gas at STP?

Chemistry

1 answer:

ICE Princess25 [194]2 years ago

7 0

Answer:

Mass = 40.4 g

Explanation:

Given data:

Mass in gram = ?

Volume of SO₂ = 14.2 L

Temperature = standard = 273 K

Pressure = standard = 1 atm

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K

T = temperature in kelvin

1 atm × 14.2 L = n × 0.0821 atm.L/ mol.K × 273 K

14.2 atm.L = n × 22.41 atm.L/ mol

n = 14.2 atm.L/22.41 atm.L/ mol

n = 0.63 mol

Mass of sulfur dioxide:

Mass = number of moles × molar mass

Mass = 0.63 mol × 64.1 g/mol

Mass = 40.4 g

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2 years ago

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The activation energy for proline isomerization of a peptide depends on the identity of the preceding residue and obeys Arrheniu

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Answer:

Activation energy of phenylalanine-proline peptide is 66 kJ/mol.

Explanation:

According to Arrhenius equation- $k=Ae^{\frac{-E_{a}}{RT}}$ , where k is rate constant, A is pre-exponential factor, $E_{a}$ is activation energy, R is gas constant and T is temperature in kelvin scale.

As A is identical for both peptide therefore-

$\frac{k_{ala-pro}}{k_{phe-pro}}=e^\frac{[E_{a}^{phe-pro}-E_{a}^{ala-pro}]}{RT}$

Here $\frac{k_{ala-pro}}{k_{phe-pro}}=\frac{0.05}{0.005}$ , T = 298 K , R = 8.314 J/(mol.K) and $E_{a}^{ala-pro}=60kJ/mol$

So, $\frac{0.05}{0.005}=e^{\frac{[E_{a}^{phe-pro}-(60000J/mol)]}{8.314J.mol^{-1}.K^{-1}\times 298K}}$

$\Rightarrow$ $E_{a}^{phe-pro}=65705J/mol=66kJ/mol$ (rounded off to two significant digit)

So, activation energy of phenylalanine-proline peptide is 66 kJ/mol

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2 years ago