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ycow [4]
3 years ago
7

Assuming all volume measurements are made at the same temperature and pressure, what volume of hydrogen gas is needed to react c

ompletely with 4.55 L of oxygen gas to produce water vapor?
Chemistry
1 answer:
Leno4ka [110]3 years ago
6 0

Answer:

9.1L

Explanation:

Firstly, to solve the problem, we need a balanced chemical equation.

2H2 + O2 ———> 2H2O

2 moles of hydrogen reacted one mole of oxygen. Now, we know that at STP, one mole of a gas occupies a volume of 22.4L, now let us get the number of moles in 4.55L of oxygen.

This means 4.55/22.4 = 0.203125 mole

Since in theory we have 2 moles of hydrogen reacting one mole of oxygen. Hence the number of moles of hydrogen we have is 0.203125 * 2 = 0.40625 mole

We now proceed to get the volume of hydrogen gas. Since 1 mole is 22.4L, then

0.40625 Is 0.40625 * 22.4 = 9.1L

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Answer:

Mg(NO4)2 is 180.3 g/mol

Explanation:

First find the substance formula.

Magnesium Nitrate.

Magnesium is a +2 charge.

Nitrate is a -1 charge.

So to balance the chemical formula,

We need 1 magnesium atom for every nitrate atom.

2(1) + 1(-2) = 0

So the substance formula is Mg(NO4)2.

Now find the molar mass of Mg(NO4)2.

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So you do this: 24.3 + 14.0(2) + 16.0(8) = 180.3 g/mol

So the molar is mass is 180.3 g/mol.

The final answer is Mg(NO4)2 is 180.3 g/mol

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