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dezoksy [38]
4 years ago
12

An automobile tire at 30°C has a pressure of 3.00 atm. Temperature decreases to -5°C. Assume that there is no volume change in t

he tire.
What is the new pressure of the tire after the temperature change?
Chemistry
1 answer:
dmitriy555 [2]4 years ago
7 0

Answer:

P₂ = 2.7 atm

Explanation:

Given data:

Initial temperature = 30°C

Initial pressure = 3.00 atm

Final temperature = -5°C

Final pressure = ?

Solution:

Initial temperature = 30°C = 30 + 273 = 303 K

Final temperature = -5°C = -5 + 273 = 268 K

According to Gay-Lussac Law,

The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.

Mathematical relationship:

P₁/T₁ = P₂/T₂

Now we will put the values in formula:

3.0 atm / 303 K = P₂/268 K

P₂ = 3.0 atm × 268 K / 303 K

P₂ = 804 atm. K /293 K

P₂ = 2.7 atm

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The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at
seraphim [82]

Answer:

the solubility of CaCO3 is 0.015g/l 25 °C

is favored at equilibrium

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The Ksp of calcium carbonate in water at 25 °C is 2.25 x 10-8. CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq) What is favored at equilibrium?

solubility is the property of a solute to dissolve in a solvent(liquid, gas ) to form a solution(soution can be saturated ,unsaturated, or supersaturated)

CaCO3(s) <----> Ca2+ (aq) + CO3 2- (aq)

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2.25x 10^-8=Ca^{2+} +CO^{2-}_{3} }

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2.25x10^-8=S*S

S^2=2.25x10^-8

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