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Varvara68 [4.7K]
3 years ago
8

Fill in the left side of this equilibrium constant equation for the reaction of nitrous acid (HNO_2) with water.

Chemistry
2 answers:
koban [17]3 years ago
6 0

Answer:

1. The equilibrium equation for the reaction between nitrous acid (HNO2) and water (H2O) is given below:

HNO2(aq) + H2O(l) <=> H3O+(aq) + NO2- (aq)

2. Ka = [H3O+] [NO2-] / [HNO2]

Explanation:

1. Step 1:

The reaction of nitrous acid HNO2 acid with water.

When nitrous acid react with water, it will produce hydronium ion (H3O+) and the corresponding nitrite ion (NO2-). This is so because an acid will dissolves in water to produce hydronium ion as the only positive ion. The equation is illustrated below:

HNO2(aq) + H2O(l) <=> H3O+(aq) + NO2- (aq)

2. Step 2:

Writing the Ka for the above equation. This is illustrated below:

The Ka is given by:

Ka = [H3O+] [NO2-] / [HNO2]

mafiozo [28]3 years ago
5 0

Answer:

Ka=\frac{[H^+][NO_2^-]}{[HNO_2]}

Explanation:

The reaction

The reaction of nitrous acid with water is a dissociation reaction of an acid:

HNO_2(aq) + H_2O(l) \longrightarrow H^+(aq) + NO_2^-(aq) + H_2O(l)

HNO_2(aq) \longrightarrow H^+(aq) + NO_2^-(aq)

The Ka equation

Ka=\frac{[H^+][NO_2^-]}{[HNO_2]}

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Answer:

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Explanation:

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7 0
3 years ago
Describe the difference between a ball-and-stick model and a space-filling model of a compound.
Lisa [10]
A space-filling model shows the relative amount of space each atom takes up. In other words, a space-filling model can show relative sizes of atoms. However, unlike ball-and-stick or structural models, space-filling models do not show bond lengths clearly. Bonds are not really like sticks in a ball-and-stick model.
4 0
4 years ago
If a sample containing 18.1 g of NH3 is reacted with 90.4 g of
USPshnik [31]

Answer:

3.64g

Explanation:

Given parameters:

Mass of NH₃  = 18.1g

Mass of Cu₂O  = 90.4g

Unknown:

Limiting reactant  = ?

Mass of N₂ formed  = ?

Solution:

The reaction equation is given as:

       Cu₂O + 2NH₃ → 6Cu + N₂ + 3H₂O

The limiting reactant is the one in short supply in the reaction. Let us find the number of moles of the given species;

  Number of moles = \frac{mass}{molar mass}  

Molar mass of Cu₂O = 2(63.6) + 16  = 143.2g/mol

Molar mass of NH₃  = 14 + 3(1) = 17g/mol

Number of moles of Cu₂O = \frac{18.1}{143.2}   = 0.13moles

Number of moles of NH₃   = \frac{90.4}{17}   = 5.32moles

  From this reaction;

       1 mole of  Cu₂O combines with 2 mole of NH₃

So   0.13moles of  Cu₂O will combine with 0.13 x 2 mole of NH₃

                                              = 0.26moles of NH₃

Therefore, Cu₂O is the limiting reactant. Ammonia is in excess;

Mass of N₂;

   Mass = number of moles x molar mass

    1 mole of Cu₂O  will produce 1 mole of N₂

    0.13 mole of Cu₂O  will produce 0.13 mole of N₂

    Mass  = 0.13 x (2 x 14) = 3.64g

5 0
3 years ago
What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 4
Ivenika [448]

Answer:

130ml of HCl(36%) in 4.90L solution => pH = 1.50

Explanation:

Need 4.90L of HCl(aq) solution with pH = 1.5.

Given pH = 1.5 => [H⁺] = 10⁻¹·⁵M = 0.032M in H⁺

[HCl(36%)] ≅ 12M in HCl

(M·V)concentrate = (M·V)diluted

12M·V(conc) = 0.032M·4.91L

=> V(conc) needed = [(0.032)(4.91)/12]Liters = 0.0130Liters or 130 ml.

Mixing Caution => Add 131 ml of HCl(36%) into a small quantity of water (~500ml) then dilute to the mark.

5 0
3 years ago
How many liters of oxygen gas, at standard temperature and pressure, will react with 25.0 grams of magnesium metal? Show all of
love history [14]
Given:
2 Mg + O2 → 2 MgO 
So,
<span>(25.0 g of Mg / 24.3051 g/mole) x (1/2) x (22.4 L/mole) = 11.5 L  of Oxygen will react with 25 grams of magnesium metal.</span>
8 0
3 years ago
Read 2 more answers
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