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3 years ago

Name one object that would be considered a gas. Describe how the atoms move in a gas.

Chemistry

2 answers:

soldi70 [24.7K]3 years ago

6 0

Answer:

gas vibrate and move freely at high speeds. liquid vibrate, move about, and slide past each other. solid vibrate (jiggle) but generally do not move from place to place.

Explanation:

Ksivusya [100]3 years ago

6 0

Answer:

Oxygen

Explanation:

The atoms in a gas move freely.

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Fossil fuels about into Earth ice age time

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4 years ago

Air pressure decreases more rapidly with altitude within a column of ________ air than in a column of __________ air.

Anarel [89]

Air Pressure drops more rapidly with altitude in a column of cold air than in warm air.The answers to this question are cold air and warm air, respectively.
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3 years ago

A compound is found to contain 13.65 % carbon and 86.35 % fluorine by mass. To answer the question, enter the elements in the or

Nimfa-mama [501]

Answer: The molecular formula will be $CF_4$

Explanation:

If percentage are given then we are taking total mass is 100 grams.

So, the mass of each element is equal to the percentage given.

Mass of C= 13.65 g

Mass of F = 86.35 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{13.65g}{12g/mole}=1.138moles$

Moles of F= $\frac{\text{ given mass of F}}{\text{ molar mass of F}}= \frac{86.35g}{19g/mole}=4.545moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{1.138}{1.138}=1$

For F = $\frac{4.545}{1.138}=4$

The ratio of C : F= 1: 4

Hence the empirical formula is $CF_4$

The empirical weight of $CF$ = 1(12)+4(19)= 88g.

The molecular weight = 88.01 g/mole

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{88.01}{88}=$

The molecular formula will be= $1\times CF_4=CF_4$

5 0

4 years ago

A certain substance has a heat of vaporization of 70.83 kJ / mol. 70.83 kJ/mol. At what Kelvin temperature will the vapor pressu

lana66690 [7]

Answer:

The answer to the question is

The temperature at which the vapor pressure will be 5.00 times higher than it was at 331 K is 353.0797 K.

Explanation:

To solve the question, we make use of the Clausius-Clapeyron equation as follows

$ln(\frac{P_2}{P_1}) = \frac{d_{vap}H}{R} (\frac{1}{T_1} -\frac{1}{T_2} )$

Where P₁ = Initial pressure

P₂ = Final pressure

T₁ = Initial temperature = 331 K

T₂ = Final temperature

dvapH = ΔvapH = Heat of vaporization = 70.83 kJ / mol.

R = Universal gas constant = 8.3145. J K⁻¹ mol⁻¹

We are required to find the temperature when P₂ = 5 × P₁

Therefore we have

$ln(\frac{5*P_1}{P_1}) = \frac{70.83}{8.3145} (\frac{1}{331} -\frac{1}{T_2} )$ = $ln(5) = 8518.853 (\frac{1}{331} -\frac{1}{T_2} )$ or T₂ = $\frac{1}{\frac{1}{331} -\frac{ln(5)}{8518.853} }$ = 353.0797 K

The vapor pressure be 5.00 times higher than it was at 331 K when the temperature is raised to 353.0797 K.

3 0

3 years ago

Compare and contrast the behavior of the food coloring and the floating pieces of paper before and after adding heat to the syst

Leviafan [203]

Answer:

the answer is c

Explanation:

7 0

3 years ago