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2 years ago

Explain the volume of ideal gasses as related to the molar concept.

Chemistry

1 answer:

love history [14]2 years ago

5 0

Answer:

n = Initial volume/22.4L

Explanation:

The molar concept is simply one that is used to find the Number of moles and explain the relationship it has with avogadro's number, molecular mass, molar mass e.t.c.

Now, in terms of molar mass, number of moles is given by the formula;

n = mass of the sample/molar mass

In terms of avogadro's number, number of moles is;

1 mole = avogadro's number = 6.02 × 10^(23)

Now, when dealing with ideal gases, the molar volume of an ideal gas is 22.4 L.

Now the relationship between this volume and the mole concept is that the number of moles is gotten by dividing the initial volume by this molar volume.

Thus;

n = Initial volume/22.4L

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Potassium chlorate (used in fireworks, flares, and safety matches) forms oxygen and potassium chloride when heated. KClO3(s) → K

Zina [86]

Answer:

10.3 g of oxygen are formed when 26.4 g of potassium chlorate is heated

Explanation:

This is the balanced equation:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

Ratio beteween the salt and oxygen is 2:3

Molar mass of KClO₃ = 122.55 g/m

Let's find out the moles of salt

Mass / Molar mass

26.4 g /122.55 g/m = 0.215 moles

So, this is the final rule of three:

If 2 moles of KClO₃ make 3 moles of oxygen

0.215 moles of KClO₃ make (0.215 .3) /2 = 0.323 moles of O₂ are produced

Molar mass O₂ = 32 g/m

Moles . molar mass = mass

0.323 m . 32g/m = 10.3 g

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3 years ago

1 What direction do<br> molecules move in?

notsponge [240]

Answer:

The majority of the molecules move from higher to lower concentration, although there will be some that move from low to high. The overall (or net) movement is thus from high to low concentration.

hope this helps!<3

3 0

2 years ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

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3 years ago

Karla added green pellets to a container of water before she began watering her lawn. What correctly describes

kvv77 [185]

The correct answer is C. The water is the solvent because the green pellets dissolved in it.

Explanation:

In solutions, the are two substances involved, the solvent and the solute. The solvent is usually a liquid substance; additionally, the solvent dissolves another substance, which is known as the solute. For example, if you dissolve a spoon of salt in a glass of water, the solute is the salt which is the substance dissolved and the solvent is the water because the solute is dissolved in it. According to this, in the case presented the water is the solvent because the green pellets which are the solute dissolve in it.

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3 years ago

Compare melting point freezing point and boiling point

Tresset [83]

Melting point- the temperature at which a substance has changed from a solid to a liquid
freezing- the temperature at which a substance chanes from liquid to a solid
boiling point- the temperature at which a substance changes from a liquid to a gas phase

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2 years ago