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3 years ago

What two properties of a gas depend on its container

Chemistry

1 answer:

katrin [286]3 years ago

6 0

uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuum i thinkitis the way its shaped

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A sealed can of your favorite soda has a carbon dioxide gas (C0) volume of 0.05 L. When it is refrigerated, the

liubo4ka [24]

Answer:

Explanation:

The new pressure can be found by using the formula for Boyle's law which is

$P_1V_1 = P_2V_2$

Since we are finding the new volume

$V_2 = \frac{P_1V_1}{P_2} \\$

From the question we have

$V_2 = \frac{207000 \times 0.05}{101000} = \frac{10350}{101000} \\ = 0.102475...$

We have the final answer as

Hope this helps you

8 0

3 years ago

Calculate the relative density of carbon monoxide and carbon dioxide with respect to air.

JulsSmile [24]

Density is proportional to molar mass, assuming pressure and temperature remain constant. Therefore, since CO has a molar mass of 28 and CO2 has a molar mass of 44:
The relative density of CO vs air is 28/29 = 0.9655.
The relative density of CO2 vs air is 44/29 = 1.517.

3 0

3 years ago

Explain why elements with less protons require fewer neutrons to be stable compared to elements with more protons.

Solnce55 [7]

Answer:

The reason for this is because of the repulsive force between protons: The stronger the repulsion force, the more neutrons are needed to stabilize the nuclei.

Explanation:

7 0

3 years ago

Predict whether the equilibria I) CH4(g) + H2O(g) ⇀↽ CO(g) + 3 H2(g), ∆H◦ = +206 kJ II) 2 SO2(g) + O2 ⇀↽ 2 SO3(g), ∆H◦ = −198 kJ

Dima020 [189]

Answer:

(1) I shifts toward product and II shifts toward reactant.

Explanation:

Increasing the temperature of an endothermic reaction (∆H is positive) shifts the equilibrium position to the right thus favoring product formation.

Increasing the temperature of an exothermic reaction (∆H is negative) shifts the equilibrium position to the left thus favoring the backward reaction.

3 0

3 years ago

What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:

Margarita [4]

Answer:

108.6 g

Explanation:

2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the <em>PV=nRT formula</em> to <u>calculate the number of nitrogen moles</u>:

P = 1.00 atm

V = 56.0 L

n = ?

R = 0.082 atm·L·mol⁻¹·K⁻¹

T = 0 °C ⇒ 0 + 273.2 = 273.2 K

<u>Inputting the data</u>:

1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K

n = 2.5 mol

Then we <u>convert 2.5 moles of N₂ into moles of NaN₃</u>, using the <em>stoichiometric coefficients of the balanced reaction</em>:

2.5 mol N₂ * $\frac{2molNaN_3}{3molN_2}$ = 1.67 mol NaN₃

Finally we <u>convert 1.67 moles of NaN₃ into grams</u>, using its <em>molar mass</em>:

1.67 mol * 65 g/mol = 108.6 g

6 0

3 years ago