If you collect 1.75 L of hydrogen gas during a lab experiment when the room temperature is 23oC and the barometric pressure is 1
1 answer:
Answer:
Explanation:
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In this case, since we can consider hydrogen gas as an ideal gas, we check the volume-pressure-temperature-mole relationship by using the ideal gas equation:
Whereas we are asked to compute the moles given the temperature in Kelvins, thr pressure in atm and volume in L as shown below:
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Answer:
Explanation:
We know we will need a balanced chemical equation with masses and molar masses, so, let's gather all the information in one place.
You don't tell us what the reaction is, but we can solve the problem so long as we balance the OH.
M_r: 58.32
Mg(OH)₂ + … ⟶ … + 2HOH
m/g: 58.3
(a) Moles of Mg(OH)₂
(b) Moles of H₂O
The molar ratio is 2 mol H₂O = 1 mol Mg(OH)₂.
The reaction will form of water.
Answer:
I know, it's hard!!
Explanation:
Answer:
52.80 % is the percent yield of the reaction.
Explanation:
Mass of nitrogen gas = 38.0 g
Moles of nitrogen =
According to reaction, 1 moles of nitrogen gas gives 2 moles of ammonia, then 2.235 moles of nitrogen will give:
ammonia
Mass of 4.470 moles of ammonia
= 4.470 mol × 17 g/mol = 75.99 g
Theoretical yield of ammonia = 217.8 g
Experimental yield of ammonia = 40.12 g
The percentage yield of reaction:
52.80 % is the percent yield of the reaction.