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den301095 [7]
3 years ago
11

There are 12 liters of nitrogen gas in a sample. How many moles of N2 in this sample?

Chemistry
1 answer:
Firlakuza [10]3 years ago
6 0

Answer:

0.54 moles of N2

Explanation:

First remember to find the volume in moles, 1 mole equals 22.4 liters.

So now use the dimensional analysis to show your work.

12 liters of N2 *  1 mol /22.4 liters of N2

Now calculate this. 12 * 1/22.4 or 12/ 22.4

12/22.4 = 0.535714286.

12 has two significant digits.

With that, The answer rounds to 0.54.

So that the final answer is 0.54 moles.

Hope it helped!

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Answer:

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Explanation:

Matter undergoes two types of changes: physical changes and chemical changes.

A physical change is one that affects only the physical properties of a substance such as colour, size, shape, etc., and in which no new substances are formed. Physical changes are easily reversible.

Chemical changes are changes which affect the chemical characteristics and properties of a substance such as reactivity, acidity, basicity, etc., resulting in the formation of new substances.

The following changes below are categorized as either physical or chemical changes based on which of the properties of the substance is affected and whether new substances result or not.

A piece of wood burns to form ash - Chemical change

Water evaporates into steam - Physical change

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The table describes how some substances were formed.
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R is the mixture formed by adding 5 g of sugar to 1 L of water.

<h3>Which is a mixture?</h3>

A mixture is an impure substance made up of two or more substances which are miscible or immiscible in each other.

Dissolving the sugar in the water makes a homogenous mixture and once dissolved a solution is made by the combination of a solute (sugar) and solvent (water).

Hence, R is the mixture formed by adding 5 g of sugar to 1 L of water.

Learn more about the mixture here:

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For the reaction 2KClO3 - 2KCI+ 302 how many moles of potassium chlorate are
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Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100-M solution of acetic
Mademuasel [1]

Answer:

1.33%

Explanation:

In an aqueous solution, a weak acid such as acetic acid, will be in equilibrium with its conjugate base, acetate ion, thus:

CH₃CO₂H(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃CO₂⁻(aq )

Where dissociation constant, ka, is defined as the ratio of concentrations of products and reactants:

Ka = 1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

<em>H₂O is not taken into account in the equilibrium because is a pure liquid</em>

<em />

When a solution of acetic acid becomes to equilibrium, the original concentration of the acid decreases producing more H₃O⁺ and CH₃CO₂⁻.

The concentrations at equilibrium when a 0.100M solution of acetic acid reaches this state, is:

[CH₃CO₂H] = 0.100M - X

[H₃O⁺] = X

[CH₃CO₂⁻] = X

<em>Where X is reaction coordinate.</em>

Replacing in Ka expression:

1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

1.8x10⁻⁵ = [X] [X] / [0.100M - X]

1.8x10⁻⁶ - 1.8x10⁻⁵X = X²

1.8x10⁻⁶ - 1.8x10⁻⁵X - X² = 0

Solving for X:

X = -0.00135 → False solution. There is no negative concentrations.

X = 0.00133 → Right solution.

That means concentration of acetate ion is:

[CH₃CO₂⁻] = 0.00133M.

Now, percent ionization is defined as 100 times the ratio between weak acid that is ionizated, [CH₃CO₂⁻] = 0.00133M, per initial concentration of the acid, [CH₃CO₂H] = 0.100M. Replacing:

% Ionization = 0.00133M / 0.100M × 100 =

<h3>1.33%</h3>

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4 0
2 years ago
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