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3 years ago

How do you differentiate between gas and a vapour ?

Chemistry

1 answer:

kolbaska11 [484]3 years ago

6 0

Gas is a substance that will not phase change.

And vapor is a substance above it's boiling point temperature.

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A gas has a volume of 3L at 200 kPa. What will its volume be if the pressure is changed to 500 kPa?

marta [7]

Answer:

Explanation:

if it's 3L per 200kPa

then it would be;

4L per 300kPa

5L per 400kPa

6L per 500kPa

that's how i'd work it out in my head, hope it helps, but not sure though!

5 0

3 years ago

In each case, calculate the appropriate ratio to show that the information given is consistent with the law of multiple proporti

Alchen [17]

Answer:

(a) 3:2; (b) 2:1

Explanation:

The Law of Multiple Proportions states that when two elements A and B combine to form two or more compounds, the masses of B that combine with a given mass of A are in the ratios of small whole numbers.

That is, if one compound has a ratio r₁ and the other has a ratio r₂, the ratio of the ratios r is in small whole numbers.

(a) Ammonia and hydrazine.

In ammonia, the mass ratio of H:N is r₁ = 0.2158/1

In hydrazine, the mass ratio of H:N is r₂ = 0.1439/1

The ratio of the ratios is:

$r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 0.2158}{0.1439} = \dfrac{1.500 }{1} = \dfrac{2.999}{2} \approx \mathbf{\dfrac{3}{2}}\\\\\text{The relative amounts of H in the two compounds are in the ratio }\boxed{\mathbf{\dfrac{3}{2}}}$

(b) Nitrogen oxides

In nitrogen monoxide, the mass ratio of O:N is r₁ = 1.142/1

In dinitrogen monoxide, the mass ratio of O:N is r₂ = 0.571/1

The ratio of the ratios is:

$r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 1.142}{0.571} = \dfrac{2.000 }{1} \approx \mathbf{\dfrac{2}{1}}\\\\\text{The relative amounts of O in the two compounds are in the ratio }\boxed{\mathbf{\dfrac{2}{1}}}$

8 0

3 years ago

Type your answer in decimal form. Do not round. 7 days = hours

djyliett [7]

168.0
Hope this helps u

5 0

2 years ago

Read 2 more answers

Plxzzzzzzzzzzzzzzzzzz

katrin [286]

The correct answers is 1/57

5 0

2 years ago

Identify the oxidizing agent and the reducing agent for 4Li(s) + O_2 (g) to 2Li_2O(s).

Doss [256]

The reaction is:

4Li(s) + O2 (g) = 2Li+ + 2O-2(s).

The oxidizing agent is the one that is being reduced which is oxygen where the charge changed from neutral to -2 while the reducing agent is the on being oxidized which is lithium where the charge change from neutral to +1.

7 0

3 years ago