To Find :
The volume of 12.1 moles hydrogen at STP.
Solution :
We know at STP, 1 mole of gas any gas occupy a volume of 22.4 L.
Let, volume of 12.1 moles of hydrogen is x.
So, x = 22.4 × 12.1 L
x = 271.04 L
Therefore, the volume of hydrogen gas at STP is 271.04 L.
Answer:
6.05g
Explanation:
The reaction is given as;
Ethane + oxygen --> Carbon dioxide + water
2C2H6 + 7O2 --> 4CO2 + 6H2O
From the reaction above;
2 mol of ethane reacts with 7 mol of oxygen.
To proceed, we have to obtain the limiting reagent,
2,71g of ethane;
Number of moles = Mass / molar mass = 2.71 / 30 = 0.0903 mol
3.8g of oxygen;
Number of moles = Mass / molar mass = 3.8 / 16 = 0.2375 mol
If 0.0903 moles of ethane was used, it would require;
2 = 7
0.0903 = x
x = 0.31605 mol of oxygen needed
This means that oxygen is our limiting reagent.
From the reaction,
7 mol of oxygen yields 4 mol of carbon dioxide
0.2375 yields x?
7 = 4
0.2375 = x
x = 0.1357
Mass = Number of moles * Molar mass = 0.1357 * 44 = 6.05g
Answer:
1) Moles of CO 2 = Given mass / Molecular mass of CO 2 = 4.4 / 44 = 0.1 mole. 2) Molecules of CO2 in 0.1 moles of CO2 = 0.1 x 6.023 x 10 23 = 6.023 x 10 22 molecules 3) 44 gram (molecular wt of CO 2) contains 2 moles atom of oxygen therefore 4.4 gram of CO2 will contain = 2 /44 *4.4 = 0.2 moles atom of Oxygen.
Answer:
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Explanation:
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