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skelet666 [1.2K]
3 years ago
12

A 4.3 L gas bubble at STP is heated to 984 K, and the pressure increases to 1520 mmHg. What is the new volume?

Chemistry
1 answer:
Natalka [10]3 years ago
5 0

Answer:

v2=40.35L

Explanation:

p1v1/t1=p2v2/t2

v2=t2p1v1/t1p2

v2=984*760*22.4/273.15*1520

v2=40.35 L

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Determine the freezing point of an aqueous solution containing 10.50 g of magnesium bromide in 200.0 g of water.
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For an aqueous solution of MgBr2, a freezing point depression occurs due to the rules of colligative properties. Since MgBr2 is an ionic compound, it acts a strong electrolyte; thus, dissociating completely in an aqueous solution. For the equation:

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ΔTf = change in freezing point = (Ti - Tf)
Ti = freezing point of pure water = 0 celsius
Tf = freezing point of water with solute = ?
Kf = freezing point depression constant = 1.86 celsius-kg/mole (for water)
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Computing for molality:
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m = 10.5g MgBr2 / 184.113/ 0.2 kg water = 0.285 mol/kg


For the problem, 
ΔTf = (Kf)(m)(i) = 1.86(0.285)(3) = 1.59 = Ti - Tf = 0 - Tf

Tf = -1.59 celsius
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