A common function of proteins and carbohydrates in the plasma membrane is...

Which of the following gases will have the highest rate of effusion at a given temperature?

Nataliya [291]

Answer:

He.

Explanation:

Graham's law states that the effusion rate is inversely proportional to the molecular weight. It means that the gas with the lowest molecular weight will have the highest effusion rate.

The molecular weights of the given gases are:

Cl₂ = 70.9 g/mol.

He = 2.0 g/mol.

O₂ = 32.0 g/mol.

Ne = 20.17 g/mol.

Since He has the smallest molecular mass, so it will have the highest rate of effusion.

5 0

3 years ago

Read 2 more answers

What is the number of grams in 3.2 moles of CuSO4×5H2O

son4ous [18]

Hey there!:

Molar mass CuSO4*5H2O = 249.68 g/mol

Therefore:

1 mole CuSO4*5H2O ---------------- 249.68 g

3.2 moles --------------------------------- ?? ( mass of CuSO4*5H2O )

mass CuSO4*5H2O = 249.68 * 3.2

mass CuSO4*5H2O = 798.97 g

Hope that helps!

7 0

3 years ago

4 points 9. Name the scientists that played a major role in the development of cell theory. *

jeyben [28]

The answer would be Theodore Schwann, Matthias Jakob Schleiden and Rudolf Virchow(he contributed but is not credited)

Hope this helps

Have a great day/night

5 0

3 years ago

Are eye drops a base or acid???

gregori [183]

base

bye hope you have a good day

3 0

4 years ago

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

3 years ago