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Naddika [18.5K]
3 years ago
14

David has a weather balloon that he fills with helium and then releases. As the balloon rises into the atmosphere, the balloon g

ets bigger and bigger. Why?
A) the temperature of the gas is decreasing, so the volume is increasing
B)the volume is increasing as the pressure inside is decreasing to match the air pressure outside the balloon
C)gas is leaking out of the balloon, so there isn't as much volume
D)this doesn't happen -- the balloon will actually stay the same size
Chemistry
1 answer:
katrin2010 [14]3 years ago
6 0

Answer: B

Explanation: since the balloon is getting bigger it would make sense for the volume to increase ( get bigger) when volume increases the pressure decreases

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What is the volume, in liters, of 1.40 mol of oxygen gas at 20.0°C and 0.974 atm?
topjm [15]

Answer:

V = 34.55 L

Explanation:

Given that,

No of moles, n = 1.4

Temperature, T = 20°C = 20 + 273 = 293 K

Pressure, P = 0.974 atm

We need to find the volume of the gas. It can be calculated using Ideal gas equation which is :

PV=nRT

R is gas constant, R=0.08206\ L-atm/mol-K

Finding for V,

V=\dfrac{nRT}{P}\\\\V=\dfrac{1.4\times 0.08206\times 293}{0.974 }\\\\V=34.55\ L

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Ana drives at a speed of 175m/s. she recorded a distance of 2km. Find the time.
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6 0
3 years ago
Read 2 more answers
Consider the reaction:
goldfiish [28.3K]

Answer:

K = Ka/Kb

Explanation:

P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?

P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka

PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb

K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)

Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)

Kb = [PCl₅]/ ([PCl₃] [Cl₂])

Since [PCl₅] = [PCl₅]

From the Ka equation,

[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)

From the Kb equation

[PCl₅] = Kb ([PCl₃] [Cl₂])

Equating them

Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])

(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)

(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)

Comparing this with the equation for the overall equilibrium constant

K = Ka/Kb

5 0
3 years ago
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