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2 years ago

Does the volume of particles affect the behavior of gas

Chemistry

1 answer:

lilavasa [31]2 years ago

7 0

Answer:

Yes, it does, although only physically and not chemically.

Explanation:

If a volume of gas is way spread out, it won't collide with the other gas particles as often, reducing pressure and temperature because they lose kinetic energy to their surroundings when they don't collide.

If it is compressed, it increases temperature and pressure because the gas particles collide with each other and the walls of the container way more often than if they had more space.

Hope this answers your question.

P.S.

Fun fact, gas particles are actually moving at 300-400 meters per second at room temperature, they only slow down to walking speed at very low temperatures, like 10 Kelvin

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Find the number of moles of water that can be formed if you have 210 mol of hydrogen gas and 100 mol of oxygen gas.

sukhopar [10]

<span>The answer is 200 mol of water. The balanced reaction is 2(H2) + (O2) = 2(H2O) The limiting reactant is O2 as it will be completely consumed first, before hydrogen gas. Hydrogen gas would need at least 105 mol oxygen gas to be consumed; in excess of the 100 mol O2. Looking at the stoichiometric coefficients, the ratio between water and oxygen is 2:1. Therefore, the water produced would be 200 moles.</span>

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2 years ago

Read 2 more answers

A 50.0 g sample of scandium, sc, is heated by exposure to 1.50 x 10 3 j. The temperature of the sc is raised by 61.1 o

statuscvo [17]

Given mass of Scandium = 50.0 g

Increase in temperature of the metal when heated = $61.1^{0}C$

Heat absorbed by Scandium = $1.50*10^{3}J$

The equation showing the relationship between heat, mass, specific heat and temperature change:

$Q = m C (deltaT)$

Where Q is heat = $1.50*10^{3}J$

m is mass = 50.0 g

ΔT = $61.1^{0}C$

On plugging in the values and solving for C(specific heat) we get,

$1.50*10^{3}J$ =50.0g(C)( $61.1^{0}C$ )

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2 years ago

Carry out the following operations as if they were calculations of experimental results and express each answer in standard nota

Shalnov [3]

Answer:

1. $10.6\; \rm m$ (one decimal place.)

2. $0.79\; \rm g$ (two decimal places.)

3. $16.5\;\rm cm^2$ (three significant figures.)

Explanation:

The first and second expressions are additions and subtractions. When adding two numbers, the accuracy of the result is given by the number of decimal places in it. The result should have as many decimal places as the input with the least number of decimal places.

For example, in the first expression:

$5.6792\;\rm m$ has four decimal places.
$0.6\; \rm m$ has only one decimal place.
$4.33\; \rm m$ has two decimal places.

Therefore, the result should be rounded to one decimal place. Note that these units are compatible for addition, since they are all the same. The result should have the same unit (that is: $\rm m$ .)

Therefore:

$\rm 5.6792\; m + 0.6\; m + 4.33\; m \approx 10.6\; \rm m$ . (Rounded to one decimal place.)

Similarly:

$\rm 3.70\; \rm g$ has two decimal places.
$2.9133\; \rm g$ has four decimal places.

Therefore, the result should be rounded to two decimal places. Its unit should be $\rm g$ (same as the unit of the two inputs.)

$\rm 3.70\; g - 2.9133\; g \approx 0.79\; \rm g$ . (Rounded to two decimal places.)

When multiplying two numbers, the accuracy of the result should be based on the number of significant figures in it. The result should have as many significant figures as the input with the least number of significant figures. In this expression:

$4.51\; \rm cm$ has three significant figures.
$3.6666\; \rm cm$ has five significant figures.

Therefore, the result should have only three significant figures.

The unit of the result is supposed to be the product of the units of the input. In this expression, that unit will be $\rm cm \cdot cm$ , which is occasionally written as $\rm cm^2$ .

$\rm 4.51 \; cm \times 3.6666 \; cm \approx 16.5\; \rm cm^2$ . (Rounded to three significant figures.)

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