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Dovator [93]
2 years ago
14

_NH, +_02 - _ NO, +_ H,0

Chemistry
1 answer:
adoni [48]2 years ago
7 0

Answer:

This equation is already balanced.

Explanation:

If your teacher requires that you put coefficients in front of each compound, then put a 1 in front of every compound.

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Determine the value of the equilibrium constant, Kgoal, for the reactionN2(g)+H2O(g)⇌NO(g)+12N2H4(g), Kgoal=?by making use of th
Black_prince [1.1K]

Answer:

K_{goal}=1.793*10^{-33}

Explanation:

N2(g)+O2(g)⇌2NO(g), K_1 = 4.10*10^{-31}

N2(g)+2H2(g)⇌N2H4(g), K_2 = 7.40*10^{-26}

2H2O(g)⇌2H2(g)+O2(g), K_3 = 1.06*10^{-10}

If we add above reaction we will get:

2N2(g)+2H2O(g)⇌2NO(g)+N2H4(g)                     Eq (1)

Equilibrium constant for Eq (1) is K_1*K_3*K_3

Divide Eq (1) by 2, it will become:

N2(g)+H2O(g)⇌NO(g)+1/2N2H4(g)                       Eq (2)  

Equilibrium constant for Eq (2) is (K_1*K_3*K_3)^{1/2}

Equilibrium constant =K_{goal}= (K_1*K_2*K_3)^{1/2}\\K_{goal}= (4.10*10^{-31} *7.40*10^{-26}*1.06*10^{-10})^{1/2}\\K_{goal}=1.793*10^{-33}

7 0
3 years ago
Which causes genetic variations and can result in different alleles? O predation rate O random mutations O competition o environ
alekssr [168]

Answer:

B

Explanation:

3 0
3 years ago
Read 2 more answers
Ik i already asked this but i need diff point of views
salantis [7]

Answer:

Is it prescribe to you?If so than yes if not then no need to

Explanation:

5 0
3 years ago
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The teacher said the volume of the liquid was 500 mL when measured a student found it was 499.7 mL what is the students percent
Natalka [10]

Answer:

<h2>0.06 % </h2>

Explanation:

The percentage error of a certain measurement can be found by using the formula

P(\%) =  \frac{error}{actual \:  \: number}  \times 100\% \\

From the question

error = 500 - 499.7 = 0.3

actual volume = 500 mL

We have

p(\%) =  \frac{0.3}{500}  \times 100 \\  =  \frac{3}{50}  \\

We have the final answer as

<h3>0.06 % </h3>

Hope this helps you

4 0
3 years ago
What is the pOH of a solution with [OH-] = 1.4 x 10-13?
lukranit [14]

Answer:

C. 12.85

Explanation:

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