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adell [148]
3 years ago
7

What is the formula for hexaboron nitride? -BN -BN6 -B6N -BN2

Chemistry
1 answer:
Softa [21]3 years ago
7 0

Answer:

Explanation:

1. Boron nitride is prepared synthetically in lab.

2. When ammonia is treated  with boric acid or boron trioxide boron nitride is prepared.This are the other reaction to produce boron nitride.

3. B  

2

​  

O  

3

​  

+2NH  

3

​  

→2BN+3H  

2

​  

O (T = 900 C)

4. B(OH)  

3

​  

+NH  

3

​  

→BN+3H  

2

​  

O (T = 900 C)

5. B  

2

​  

O  

3

​  

+CO(NH  

2

​  

)  

2

​  

→2BN+CO  

2

​  

+2H  

2

​  

O (T > 1000 C)

6. B  

2

​  

O  

3

​  

+3CaB  

6

​  

+10N  

2

​  

→20BN+3CaO (T > 1500 C)

Properties of Boron nitride :  

1. Boron nitride is  ionic in nature hence it reduces he co valency and electrical conductivity.

2. Hexa-boron nitride is thermally stable.1. Boron nitride is prepared synthetically in lab.

2. When ammonia is treated  with boric acid or boron trioxide boron nitride is prepared.This are the other reaction to produce boron nitride.

3. B  

2

​  

O  

3

​  

+2NH  

3

​  

→2BN+3H  

2

​  

O (T = 900 C)

4. B(OH)  

3

​  

+NH  

3

​  

→BN+3H  

2

​  

O (T = 900 C)

5. B  

2

​  

O  

3

​  

+CO(NH  

2

​  

)  

2

​  

→2BN+CO  

2

​  

+2H  

2

​  

O (T > 1000 C)

6. B  

2

​  

O  

3

​  

+3CaB  

6

​  

+10N  

2

​  

→20BN+3CaO (T > 1500 C)

Properties of Boron nitride :  

1. Boron nitride is  ionic in nature hence it reduces he co valency and electrical conductivity.

2. Hexa-boron nitride is thermally stable.

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7 0
3 years ago
If the δh°soln of hno3 is –33.3 kj/mol, then how much heat is evolved by dissolving 0.150 mol hno3 in 100.0 ml of water?
Leto [7]
Answer: - 5.00 kJ

Explanation:

1) The heat evolved during the reation is proportional to the number of moles dissolved.

2) Proportionality = ratio is constant

<span>3) δh°soln of HNO3 is –33.3 kJ/mol => ratio 1

                  -33.3 kJ
ratio 1 = ---------------------
                 1 mol HNO3

4) ratio 2

        x
--------------
0.150 mol     
</span>
5) proportion

ratio 1 = ratio 2

        x                -33.3 kJ
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4 0
3 years ago
For the reaction: 3 H2(g) + N2(g) &lt;--&gt; 2 NH3(g), the concentrations at equilbrium were [H2] = 0.10 M, [N2] = 0.10 M, and [
masya89 [10]

The equilibrium constant, k of the reaction in which case, the concentrations of the given reactants and products are as indicated is; Choice A; K = 3.1 x 10⁵

<h3>What is the equilibrium constant , k of the reaction as described in the task content?</h3>

It follows from above that the concentrations of the reactants and products are as follows; [H2] = 0.10 M, [N2] = 0.10 M, and [NH3] = 5.6 M at equilibrium.

Hence, the equilibrium constant of the reaction in discuss is;

K = [5.6]²/[0.10]³[0.10]

k = 5.6² × 10⁴

k = 3.136 × 10⁵

K = 3.1 × 10⁵.

Read more on equilibrium constant;

brainly.com/question/1619133

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5 0
2 years ago
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alex41 [277]

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