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3 years ago

Which quantity is most closely related to the speed of gas particles?

Chemistry

2 answers:

just olya [345]3 years ago

5 0

Answer: Temperature

Explanation: According to Kinetic molecular Theory of gases temperature is directly proportional to the kinetic energy of the gas particle which leads to the speed of the gas.

Thus When the temperature increases, thermal motion of the particles increases which lead to the increase in the speed of the particle as they start moving in random directions.

They collide with each other and with the walls of the container during this process as well.

max2010maxim [7]3 years ago

3 0

The temperature of the gas, the hotter they get the faster they move. the colder they get, the slower they move

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If 54 g of Al reacted with 160 g of O2, find out the weight of product

Oxana [17]

Answer:

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3 years ago

Roman55 [17]

Answer:

305 litres of NO gas will be produced from 916 L of NO₂

Explanation:

Given the balanced equation of the chemical reaction as follows:

3 NO₂ (g) + H₂O( l) —— 2 HNO₃ (l) + NO (g)

Under standard conditions, 3 moles of No₂ will react with 1 mole of water to produce 1 mole of NO gas.

Molar volume of a gas at STP is 22.4 L

Number of moles of NO₂ gas present in 916 L = 916/22.4 = 40.893 moles of NO₂ gas

From the mole ratio of NO₂ to NO in the equation of reaction,

Number of moles of NO that will be produced = 1/3 × 40.893 moles = 13.631 moles of NO gas

Volume of 13.631 moles of NO gas = 13.631 × 22.4

Volume of NO gas produced = 305.334L

Therefore, Volume of NO gas produced from the reaction of 916 L of NO₂ with water = 305 L

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2 years ago

I NEED HELP PLEASE, THANKS! :)

marin [14]

Answer:

$\large \boxed{1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2 }}$

Explanation:

1. Calculate the moles of copper(II) hydroxide

$\text{Moles of Cu(OH)}_{2} = \text{23.45 g Cu(OH)}_{2} \times \dfrac{\text{1 mol Cu(OH)}_{2}}{\text{97.562 g Cu(OH)}_{2}} = \\\\\text{0.240 36 mol Cu(OH)}_{2}$

2. Calculate the molecules of copper(II) hydroxide

$\text{No. of molecules} = \text{0.240 36 mol Cu(OH)}_{2} \times \dfrac{6.022 \times 10^{23}\text{ molecules Cu(OH)}_{2}}{\text{1 mol Cu(OH)}_{2}}\\\\= 1.447 \times 10^{23}\text{ molecules Cu(OH)}_{2}\\\text{The sample contains $\large \boxed{\mathbf{1.447 \times 10^{23}}\textbf{ molecules Cu(OH)}_{\mathbf{2}}}$}$

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Explanation:

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