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KengaRu [80]
3 years ago
12

What mass, in grams, of oxygen gas (o2) is contained in a 8.5-liter tank at 32.0 degrees celsius and 3.23 atmospheres?

Chemistry
1 answer:
e-lub [12.9K]3 years ago
5 0
For this problem we assume that oxygen is an ideal gas. So, we use the equation PV=nRT where P is pressure, V is the volume, n is the number of moles, R is a universal constant and T is the temperature. We first solve for the number of moles n. Then, using the molar mass of oxygen we convert it to grams. 

PV=nRT
n = PV / RT
n = 3.23 (8.5) / 0.08206 (32+273.15)
n = 1.0964 mol 

mass = 1.0964 mol (32g / 1 mol) = 35.09 g O2
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A sailor on a trans-Pacific solo voyage notices one day that if he puts 735.mL of fresh water into a plastic cup weighing 25.0g,
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Answer:

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Explanation:

According to Archimedes' principle, mass of freshwater and cup = mass of equal volume of seawater

mass of freshwater = density * volume

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mass of freshwater + cup = 734.265 + 25 = 759.265 g

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Volume of seawater displaced = 735 mL = 0.735 L (assuming the cup volume is negligible)

1 liter = 1000 cm³ = 1000 mL;

Density of seawater = mass / volume

Density of seawater = 759.265 g / 0.735 L = 1033.01 g/L

Density of freshwater in g/L = 0.999 g/ (1/1000) L = 999 g/L

mass of 1 Liter seawater = 1033.01 g

mass of 1 Liter freshwater = 999 g

mass of salt dissolved in 1 L of seawater = 1033.01 g - 999 g = 34.01 g

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