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Lemur [1.5K]
1 year ago
10

A sample of hydrogen gas was collected over water at 21ºc and 685 mmhg. the volume of the container was 7.80 l. calculate the m

ass of h2(g) collected. (vapor pressure of water = 18.6 mmhg at 21c.)
Chemistry
1 answer:
murzikaleks [220]1 year ago
8 0
To determine the mass of the hydrogen gas that was collected, we calculate for the moles of hydrogen gas from the conditions given. In order to do this, we need an equation which would relate pressure, volume and temperature. For simplicity, we assume the gas is an ideal gas so we use the equation PV = nRT where P is the pressure, V is the volume, n is the number of moles of the gas, T is the temperature and R is the universal gas constant. We calculate as follows:

PV = nRT
n = PV / RT
n = (18.6/760) (7.80) / 0.08205 ( 21 + 273.15)
n = 0.0079 mol

Mass = 0.0079 mol ( 18.02 g / mol ) = 0.1425 g H2
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2 years ago
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How many liters of a 0.225 M solution of KI are needed to contain 0.935 moles of KI?
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4.16L

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2 years ago
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You find a little bit (0.150g) of a chemical marked Tri-Nitro-Toluene, and upon combusting it in oxygen, collect 0.204 g of CO2
avanturin [10]

1) Answer is: the formula is C₇N₃O₆H₅.

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ω(N) = 18.5% ÷ 100%.

ω(N) = 0.185; mass percentage of the nitrogen.

m(N) = 0.150 g · 0.185.

m(N) = 0.02775 ·; mass of the nitrogen.

n(N) = 0.02775 g ÷ 14 g/mol.

n(N) = 0.002 mol; amount of the nitrogen.

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n(CO₂) = 0.0046 mol.

n(C) = n(CO₂) = 0.0046 mol; amount of the carbon.

m(C) = 0.0046 mol · 12 g/mol.

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n(H₂O) = 0.00166 mol.

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m(H) = 0.0033 mol · 1 g/mol.

m(H) = 0.0033 g; mass of the hydrogen.

2) m(O) = m(TNT) - m(N) - m(C) - m(H).  

m(O) = 0.150 g - 0.02775 g - 0.0552 g - 0.0033 g.

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n(O) = 0.06375 g ÷ 16 g/mol.

n(O) = 0.004 mol; amount of oxygen.

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n(C) : n(N) : n(O) : n(H) = 2.33 : 1 : 2 : 1.66 /×3.

n(C) : n(N) : n(O) : n(H) = 7 : 3 : 6 : 5.

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