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Bogdan [553]
2 years ago
7

The volume of 1.20 lol of gas at 61.3 kPa and 25.0 degree Celsius is

Chemistry
1 answer:
aniked [119]2 years ago
4 0

Answer:

V = 48.5 L

Explanation:

Converting °C to K and kPa to atm

T = 25.0°C + 273.15 = 298.15 K

P = 61.3 kPa × (1 atm / 101.325 kPa) = 0.60498 atm

Calculating the volume of gas

V = nRT / P

V = (1.20 mol)(0.082057 L•atm/mol•K)(298.15 K) / 0.60498 atm

V = 48.5 L

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Answer:

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8 0
3 years ago
An enzyme with molecular weight of 310 kDa undergoes a change in shape when the substrate binds. This change can be characterize
oee [108]

Answer:

(a) r = 6.26 * 10⁻⁷cm

(b) r₂ = 6.05 * 10⁻⁷cm

Explanation:

Using the sedimentation coefficient formula;

s =  M(1-Vρ) / Nf ; where s is sedimentation coefficient, M is molecular weight, V is specific volume of protein, p is density of the solvent, N is Avogadro number, f if frictional force = 6πnr, n is viscosity of the medium, r is radius of particle

s = M ( 1 - Vρ) / N*6πnr

making r sbjct of formula, r =  M (1 - Vρ) / N*6πnrs

Note: S = 10⁻¹³ sec, 1 KDalton = 1 *10³ g/mol, I cP = 0.01 g/cm/s

r = {(3.1 * 10⁵ g/mol)(1 - (0.732 cm³/g)(1 g/cm³)} / { (6.02 * 10²³)(6π)(0.01 g/cm/s)(11.7 * 10⁻¹³ sec)

r = 6.26 * 10⁻⁷cm

b. Using the formula r₂/r₁ = s₁/s₂

s₂ = 0.035 + 1s₁ = 1.035s₁

making r₂ subject of formula; r₂ = (s₁ * r₁) / s₂ = (s₁ * r₁) / 1.035s₁

r₂ = 6.3 * 10⁻⁷cm / 1.035

r₂ = 6.05 * 10⁻⁷cm

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