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3 years ago

There state of matter that keeps its shape and volume when it is placed in a container

Chemistry

1 answer:

Lostsunrise [7]3 years ago

3 0

It'd liquid because it's take up space and it doesn't have a definite shape

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A cylinder is filled with 10.0L of gas and a piston is put into it. The initial pressure of the gas is measured to be 209.kPa. T

disa [49]

Answer : The final pressure of the gas will be, 26.8 kPa

Explanation :

According to the Boyle's law, the pressure of the gas is inversely proportional to the volume of the gas at constant temperature of the gas and the number of moles of gas.

$P\propto \frac{1}{V}$

or,

$PV=k$

or,

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of the gas = 209 kPa

$P_2$ = final pressure of the gas = ?

$V_1$ = initial volume of the gas = 10.0 L

$V_2$ = final volume of the gas = 78.0 L

Now put all the given values in this formula, we get the final pressure of the gas.

$209kPa\times 10.0L=P_2\times 78.0L$

$P_2=26.8kPa$

Therefore, the final pressure of the gas will be, 26.8 kPa

3 0

3 years ago

Methane reacts with oxygen in the following combustion reaction.

Natali5045456 [20]

The methane molecule undergoes oxidation and the carbon is oxidized to carbon dioxide and hydrogen to water.

In this reaction there is cleavage of four C-H bonds and two O=O bonds

there will be formation of two C=O bonds and four O-H bonds.

Overall due to more bond cleavage energy there will be evolution of energy. thus combustion is an exothermic reaction.

6 0

3 years ago

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Example of a neutral mutation?

Norma-Jean [14]

The answer is D

Explanation: A RED FLOWER WITH ONE PART COLORED YELLOW

3 0

3 years ago

Read 2 more answers

Balance :FeCl3 + __Ca(OH)2 → ___ Fe(OH)3 + __ CaCl2

ArbitrLikvidat [17]

Your answer is 2:3:2:3

7 0

3 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