What does the kinetic theory state about the relationship between the speed and temperature of gas molecules?

1 answer:

5 0

The kinetic theory state about the relationship between the speed and temperature of gas molecules "As the temperature increases, the speed of gas molecules increases".

Option: C

Explanation:

The temperature relates to a gas, according to the kinetic theory of gases, influences the rate of motion of the gas particles. The higher the gas temperature, the faster the gas particles are moving; the reverse is true, too.

This is because the gas particles gain more kinetic energy as the temperature of the gas increases, and therefore move faster and collide more with each other and the container walls.The rate of a gas's molecules is proportional to the temperature, and is inversely proportional to the gas's molar mass.

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The number of grams of H2 in 1470 mL of H2 gas.

Lorico [155]

Answer:

0.1313 g.

Explanation:

It is known that at STP, 1.0 mole of ideal gas occupies 22.4 L.

Suppose that hydrogen behaves ideally and at STP conditions.

Using cross multiplication:

1.0 mol of hydrogen occupies → 22.4 L.

??? mol of hydrogen occupies → 1.47 L.

∴ The no. of moles of hydrogen that occupies 1.47 L = (1.0 mol)(1.47 L)/(22.4 L) = 6.563 x 10⁻² mol.

Now, we can get the no. of grams of hydrogen in 6.563 x 10⁻² mol:

The no. of grams of hydrogen = no. of hydrogen moles x molar mass of hydrogen = (6.563 x 10⁻² mol)(2.0 g/mol) = 0.1313 g.

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Calculate δ h for the reaction:no (g) + o2 (g) ↔ no2 (g). given: 2o3(g) ↔ 3o2(g) δh=-426 kj o2(g) ↔ 2o(g) δh=+ 490 kj no(g) + o3

maks197457 [2]

To calculate the δ h, we must balance first the reaction:

NO + 0.5O2 -----> NO2

Then we write all the reactions,

2O3 -----> 3O2 δ h = -426 kj eq. (1)

O2 -----> 2O δ h = 490 kj eq. (2)

NO + O3 -----> NO2 + O2 δ h = -200 kj eq. (3)

We divide eq. (1) by 2, we get

O3 -----> 1.5O2 δ h = -213 kj eq. (4)

Then, we subtract eq. (3) by eq. (4)

NO + O3 -----> NO2 + O2 δ h = -200 kj
- (O3 -----> 1.5 O2 δ h = -213 kj)
NO -----> NO2 - 0.5O2 δ h = 13 kj eq. (5)

eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O -----> 0.5O2 δ h = -245 kj eq. (6)

Add eq. (6) to eq. (5), we get

NO -----> NO2 - 0.5O2 δ h = 13 kj
+ O -----> 0.5O2 δ h = -245 kj
NO + O ----> NO2 δ h = -232 kj

ANSWER: NO + O ----> NO2 δ h = -232 kj

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Answer:

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