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Nady [450]
3 years ago
15

I need help on this and the first person who answer correctly gets a BRANLIST..

Chemistry
1 answer:
stealth61 [152]3 years ago
4 0

Answer:

1 oxygen

Explanation:

cause H2O= two hydrogen + one oxyge

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What is the molar mass of Agl2​
Maru [420]

Answer:

107.8682

Explanation:

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7 0
3 years ago
2 H2S(g) ⇄ 2 H2(g) + S2(g) Kc = 9.3× 10-8 at 400ºC 0.47 moles of H2S are placed in a 3.0 L container and the system is allowed t
Anon25 [30]

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 1.648\times 10^{-3}M

<u>Explanation:</u>

We are given:

Initial moles of hydrogen sulfide gas = 0.47 moles

Volume of the container = 3.0 L

The molarity of solution is calculated by using the equation:

\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of the solution}}

So, \text{Initial molarity of hydrogen sulfide gas}=\frac{0.47}{3}=0.1567M

The given chemical equation follows:

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

<u>Initial:</u>                  0.1567

<u>At eqllm:</u>           0.1567-2x       2x         x

The expression of K_c for above equation follows:

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

We are given:

K_c=9.3\times 10^{-8}

Putting values in above equation, we get:

9.3\times 10^{-8}=\frac{(2x)^2\times x}{(0.1567-2x)^2}\\\\x=8.24\times 10^{-4}

So, equilibrium concentration of hydrogen gas = 2x=(2\times 8.24\times 10^{-4})=1.648\times 10^{-3}M

Hence, the concentration of hydrogen gas at equilibrium is 1.648\times 10^{-3}M

7 0
3 years ago
Another metal phosphate is cobalt phosphate. It will behave similar to calcium phosphate in an acid solution. What is the net io
ryzh [129]
We are given with
Cobalt phosphate - CoPO4

We are asked for the net ionic equation for the phosphate dissolving in H3O+

The net ionic equation is
CoPO4 (s) + H3O+ (aq) ----->  HPO42- (aq) + Co3+ (aq) + H2O *(l)
8 0
3 years ago
Read 2 more answers
Y-2=(x+3) I don't know how to set it up to solve
Alinara [238K]
Y EQUALS X - 1 I MIGHT BE WRONG

5 0
4 years ago
The combustion of ammonia in the presence of excess oxygen yields no2 and h2o: 4 nh3 (g) + 7 o2 (g) → 4 no2 (g) + 6 h2o (g) the
marshall27 [118]
The combustion of ammonia in presence of excess oxygen yields NO2 and H2O.
The molar mass of ammonia is 17.02 g/mol
Therefore, moles of ammonia in 43.9 g
          = 43.9 /17.02
          = 2.579 moles
From the equation the mole ratio of ammonia to nitrogen iv oxide is 4:4
The molar mass of NO2 is 46 g/mol
The number of moles of NO2 is the same as that of ammonia since they have equal ratio,
 = 2.579 moles
Therefore, mass of NO2
   = 2.579 moles ×46
   = 118.634 g
   ≈ 119 g


3 0
4 years ago
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