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Hunter-Best [27]
3 years ago
9

Dinitrogen oxide (N₂O) gas was generated from the thermal decomposition of ammonium nitrate and collected over water. The wet ga

s occupied 130 mL at 21◦C when the atmospheric pressure was 750 Torr. What volume would the same amount of dry dinitrogen oxide have occupied if collected at 750 Torr and 21 ◦C? The vapor pressure of water is 18.65 Torr at 21◦C. Answer in units of mL.
Chemistry
1 answer:
RideAnS [48]3 years ago
4 0

Answer:

126.73 mL

Explanation:

The total pressure of the gas mixture is the sum of the vapor pressure of its constituents. So, the vapor pressure of N₂O(p) can be calculated:

750 = 18.85 + p

p = 750 - 18.85

p = 731.15 torr

It means that for 731.15 torr, N₂O occupied 130 mL. For the general gas equation, we know that

\frac{p1V1}{T1} = \frac{p2V2}{T2}

Where <em>p</em> is the pressure, <em>V</em> is the volume, <em>T</em> is the temperature, 1 is the initial state, and 2 the final state. For the same temperatue (21ºC), the equation results on Boyle's law:

p1V1 = p2V2, so:

731.15x130 = 750xV2

750V2 = 95049.5

V2 = 126.73 mL

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1)Explain how different observations and experiments led to changes in the atomic model.
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6 0
3 years ago
Which element is a nonmetallic liquid at room temperature
fgiga [73]

Answer:

water!

Explanation:

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6 0
2 years ago
Consider the reaction 4Fe(s) + 3O2(g) --&gt; 2Fe2O3(s) How many grams of oxygen are needed to produce 12.5 g of Fe2O3 (iron(III)
vlabodo [156]

Answer:

3.76 g of O₂ are needed to produced 12.5 g of Fe₂O₃

Explanation:

The reaction is:  4Fe (s) + 3O₂ (g)  →  2Fe₂O₃ (s)

4 moles of iron react to 3 moles of oxygen in order to produce 2 moles of iron (III) oxide.

Let's determine the moles of the produced product.

12.5 g . 1mol/ 159.69g = 0.0783 moles

If we assume Iron in excess, we work with the oxygen.

2 moles of Fe₂O₃ are produced by 3 moles of oxygen

Then, 0.0783 moles of Fe₂O₃ might be produced by (0.0783 . 3)/2

0.117 moles.

We convert the moles to mass → 0.117 mol . 32 g/1mol = 3.76 g

3 0
3 years ago
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