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Bas_tet [7]
3 years ago
12

Which of the following apply to gases. Select all that apply. Gas collisions are elastic. Gases mix faster than solids or liquid

s. Gases with larger molecular weights diffuse slower than gases with lower molecular weights. Gases with larger molecular weights diffuse faster than gases with lower molecular weights. Gas particles can move like solids and liquids along with translational motion.
Chemistry
2 answers:
PSYCHO15rus [73]3 years ago
7 0

Answer:

-Gas collisions are elastic

- Gases mix faster than solids or liquids-

Gases with larger molecular weights diffuse slower than gases with lower molecular weights.

Explanation:

Matter is any substance that has weight and occupy space. There are three classes of matter. Theses are solid, liquid and gas.

Under Kinetic Theory of Gas

1. Gases move randomly in a straight line, thereby colliding with themselves and the walls of their containers

2. The collision of a gas molecule are perfectly elastic

3. temperature is a measure of average kinetic energy

Bearing this at the back of our minds, we can say that the following from the option applies to gases

-Gas collisions are elastic

- Gases mix faster than solids or liquids-

Gases with larger molecular weights diffuse slower than gases with lower molecular weights.

And these are not true

Gases with larger molecular weights diffuse faster than gases with lower molecular weights. Gas particles can move like solids and liquids along with translational motion.

marishachu [46]3 years ago
5 0
<span> Gas collisions are elastic. Gases mix faster than solids or liquids. Gases with larger molecular weights diffuse slower than gases with lower molecular weights. </span>
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If 175mL of oxygen is produced at STP, how many grams of hydrogen peroxide, H2O2
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Answer:

0.53g

Explanation:

We'll begin by converting 175mL to L. This is illustrated below:

1000mL = 1L

Therefore 175mL = 175/1000 = 0.175L

Next, we shall calculate the number of mole of O2 that occupy 0.175L. This is illustrated below:

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Therefore, Xmol of O2 will occupy 0.175L i.e

Xmol of O2 = 0.175/22.4

Xmol of O2 = 7.81×10¯³ mole

Therefore, 7.81×10¯³ mole of O2 occupy 175mL.

Next, we shall determine the number of mole of H2O2 that decomposed to produce 7.81×10¯³ mole of O2. This is illustrated below:

2H2O2 —> 2H2O + O2

From the balanced equation above,

2 moles of H2O2 decomposed to produce 1 mole of O2.

Therefore, Xmol of H2O2 will decompose to produce 7.81×10¯³ mole of O2 i.e

Xmol of H2O2 = 2 x 7.81×10¯³

Xmol of H2O2 = 1.562×10¯² mole

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Molar mass of H2O2 = (2x1) + (16x2) = 34g/mol

Mole of H2O2 = 1.562×10¯² mole

Mass of H2O2 =..?

Mole = mass /Molar mass

1.562×10¯² = mass /34

Cross multiply

Mass of H2O2 = 1.562×10¯² x 34

Mass of H2O2 = 0.53g

Therefore, 0.53g of Hydrogen peroxide, H2O2 were decomposition in the reaction.

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