What would improve Hanson's experiment would be if he:
A. If he measured the volume of the soil beforehand
A. If he measured the volume of the soil beforehand
A mass of 10 kg has a velocity of 123 m/s to the north. What is the momentum
1 answer:
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Answer:
(a) Oxygen
(b) 0.84 g
(c) 2.54 g
Explanation:
(a)
Explanation:
The formula for the calculation of moles is shown below:
Given: For butane
Given mass = 1.00 g
Molar mass of butane = 58.0 g/mol
Moles of butane = 1.00 g / 58.0 g/mol = 0.0172 moles
Given: For
Given mass = 3.00 g
Molar mass of = 32.0 g/mol
Moles of = 3.00 g / 32.0 g/mol = 0.09375 moles
According to the given reaction:
2 moles of butane react with 13 moles of
1 mole of butane react with 13/2 moles of
0.0172 moles of butane react with (13/2)*0.0172 moles of
Moles of required = 0.1118 moles
Available moles of = 0.09375 moles
<u>Limiting reagent is the one which is present in small amount. Thus, is limiting reagent. (0.09375 < 0.1118 )</u>
(b)
The formation of the product is governed by the limiting reagent. So,
13 moles of react with 2 moles of butane
1 mole of react with 2/13 moles of butane
0.09375 mole of react with (2/13)*0.09375 moles of butane
Moles of butane used = 0.0144 moles
Molar mass of butane = 58.0 g/mol
<u>Mass of butane used = Moles × Molar mass = 0.0144 × 58.0 g = 0.84 g</u>
(c)
13 moles of on reaction forms 8 moles of carbon dioxide
1 mole of on reaction forms 8/13 moles of carbon dioxide
0.09375 mole of on reaction forms (8/13)*0.09375 moles of carbon dioxide
Moles of carbon dioxide obtained = 0.05769 moles
Molar mass of = 44.0 g/mol
<u>Mass of = Moles × Molar mass = 0.05769 × 44.0 g = 2.54 g</u>