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2 years ago

Add a constant temperature when the volume of the gas is decreased what happens to its pressure

Chemistry

1 answer:

Margaret [11]2 years ago

8 0

Answer:

<h2>Pressure will increase</h2>

Explanation:

At a constant temperature, the pressure of gas will increase proportional to the decrease in volume of the gas.

P1V1= P2V2

Decrease in volume result in increase in pressure as the equation has to hold true.

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Two chemist are trying to determine if their water sample is a mixture or pure substance. In their investigation they first filt

Effectus [21]

The correct answer is option B. Dirty water is a mixture of solid particles and liquid. It is both a mixture and pure substance.

The dirty water sample has both gravel and liquid water in it. After filtration the gravel is removed so the water sample looks clearer than before filtration. Liquid water is a pure substance because it is a compound that is made up of elements hydrogen and oxygen. Now the gravel is only physically combined with the liquid water, thus giving the water sample properties of a mixture. And like any mixture, gravel is physically separated through filtration from the liquid water.

Thus the water sample of the chemists is both a mixture and pure substance.

8 0

3 years ago

A fixed mass of oxygen gas occupies 300cm cube at 0 degree centigrade. what volume would the gas occupy at 15 degree centigrade

egoroff_w [7]

Answer:

Volume occupied by oxygen gas at 15 degree centigrade is equal to $316.5$ centimeter cube

Explanation:

Assuming Pressure is constant.

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

where T1 and T2 are temperature in Kelvin

Substituting the give values we get-

$\frac{300}{273} = \frac{V_2}{288}$

$V_2 = \frac{288*300}{273} \\V_2 = 316.5$

Volume occupied by oxygen gas at 15 degree centigrade is equal to $316.5$ centimeter cube

5 0

2 years ago

The gas phase decomposition of sulfuryl chloride at 600 K SO2Cl2(g)SO2(g) + Cl2(g) is first order in SO2Cl2. During one experime

krok68 [10]

Answer: The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

Explanation:

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant

t = age of sample = 559 min

a = let initial amount of the reactant = $2.83\times 10^{-3}$

a - x = amount left after decay process = $3.06\times 10^{-4}$

$559min=\frac{2.303}{k}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=\frac{2.303}{559}\log\frac{2.83\times 10^{-3}}{3.06\times 10^{-4}}$

$k=3.96\times 10^{-3}min^{-1}$

The rate constant for the reaction is $3.96\times 10^{-3}min^{-1}$

6 0

3 years ago

How do the independent and dependent variables in an experiment compare?

DENIUS [597]

An independent variable<span> is the </span>variable<span> that is changed or controlled in a scientific </span>experiment<span> to test the effects on the </span>dependent variable<span>. </span>

4 0

3 years ago

Solids or liquids are not included in an equilibrium expression because:

Harlamova29_29 [7]

I believe the answer is D

7 0

3 years ago