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iris [78.8K]
2 years ago
12

5. __NH3 + __O2 >>>>>__ NO +__ H20

Chemistry
1 answer:
inn [45]2 years ago
5 0
2 NH3+ 2 O2 —> 2 NO+ 3 H2O
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bonufazy [111]

Answer:

The law of conservation of energy states that the total energy is constant in any process. Energy may change in form or be transferred from one system to another, but the total remains the same.

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8 0
2 years ago
For the equilibrium
Mamont248 [21]

Answer:

Equilibrium concentrations of the gases are

H_2S=0.596M

H_2=0.004 M

S_2=0.002 M

Explanation:

We are given that  for the equilibrium

2H_2S\rightleftharpoons 2H_2(g)+S_2(g)

k_c=9.0\times 10^{-8}

Temperature, T=700^{\circ}C

Initial concentration of

H_2S=0.30M

H_2=0.30 M

S_2=0.150 M

We have to find the equilibrium concentration of gases.

After certain time

2x number of moles  of reactant reduced and form product

Concentration of

H_2S=0.30+2x

H_2=0.30-2x

S_2=0.150-x

At equilibrium

Equilibrium constant

K_c=\frac{product}{Reactant}=\frac{[H_2]^2[S_2]}{[H_2S]^2}

Substitute the values

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

9\times 10^{-8}=\frac{(0.30-2x)^2(0.150-x)}{(0.30+2x)^2}

By solving we get

x\approx 0.148

Now, equilibrium concentration  of gases

H_2S=0.30+2(0.148)=0.596M

H_2=0.30-2(0.148)=0.004 M

S_2=0.150-0.148=0.002 M

3 0
2 years ago
6. What are two uses for nonmetals?
lesantik [10]
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5 0
2 years ago
Hydrogen gas explosive within range of 4%-75% v/v. Assuming that each student in your class produces 6 L of H2 and that the lab
SpyIntel [72]
To determine whether the amount of H2 in the lab is dangerous, we first need to know how much hydrogen gas is present in the room in units of percent by volume. For this particular problem, we cannot exactly determine since we do not know the total volume of the room. Hope this answers the question.
3 0
3 years ago
Which statement is correct? A. pKa is not an indicator of acid strength. B. An acid with a small Ka is stronger than an acid wit
Svetradugi [14.3K]

Answer:

D

Explanation:

Since [pKa = - log Ka]....hence..,the larger the Ka value,the stronger the acid is..so this means that the pKa is vice versa

Saying that the smaller the pKa value..the stronger the acid is.

3 0
3 years ago
Read 2 more answers
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