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goblinko [34]
3 years ago
8

Please Help!!! 25 POINTS!!!

Chemistry
2 answers:
son4ous [18]3 years ago
5 0

Answer:

1. independent

2. question

3. same

4. true

Neporo4naja [7]3 years ago
3 0
Independent question same true
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Answer:

temperature controlled

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The pressure of a sample of helium is 1.556 atm in a 268.5 mL container. If the container is compressed to 112.4 mL without chan
Rainbow [258]

Answer:

a.  3.72 [atm]

Explanation:

For a gas at constant temperature, (with no change in number of molecules of the gas), we can apply Boyle's Law:  P_1V_1=P_2V_2

(1.556[atm])(268.5[mL])=P_2(112.4[mL])

\dfrac{(1.556[atm])(268.5[mL\!\!\!\!\!\!\!\!{--}])}{112.4[mL \!\!\!\!\!\!\!\!{--}]}=\dfrac{P_2(112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----})}{112.4[mL]\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!{-----}}

3.716957[atm]=P_2

It seems like the answer should have 4 significant figures since all of the other quantities have 4 significant figures, but the closest answer choice of those provided is a.  3.72

3 0
2 years ago
What determines the state of a substance
frez [133]

The kinetic energies of the particles (atoms, molecules, or ions) that make up a substance or object.

4 0
3 years ago
Read 2 more answers
Think about a carbon atom that is released into the atmosphere from the burning of wood in a campfire. If it were to go through
viva [34]
1) <span>Step 5 A tree absorbs the carbon from the atmosphere into its leaves for photosynthesis. Inorganic carbon is turned into organic.
2) </span><span>Step 1 A caterpillar gets the carbon by eating the tree's leaves. Caterpillar use carbon for energy.
3) </span><span>Step 3 A bird gets the carbon by eating the caterpillar. Organic carbon shifts from one animal to another
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3 0
3 years ago
What is the mean free path for the molecules in an ideal gas when the pressure is 100 kPa and the temperature is 300 K given tha
sladkih [1.3K]

Answer:

The mean free path = 2.16*10^-6 m

Explanation:

<u>Given:</u>

Pressure of gas P = 100 kPa

Temperature T = 300 K

collision cross section, σ = 2.0*10^-20 m2

Boltzmann constant, k = 1.38*10^-23 J/K

<u>To determine:</u>

The mean free path, λ

<u>Calculation:</u>

The mean free path is related to the collision cross section by the following equation:

\lambda =\frac{1}{n\sigma }------(1)

where n = number density

n = \frac{P}{kT}-----(2)

Substituting for P, k and T in equation (2) gives:

n = \frac{100,000 Pa}{1.38*10^{-23} J/K*300K} =2.42*10^{25}\  m^{3}

Next, substituting for n and σ in equation (1) gives:

\lambda =\frac{1}{2.42*10^{25}m^{-3}* .0*10^{-20}m^{2}}=2.1*10^{-6}m

6 0
3 years ago
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