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uysha [10]
2 years ago
15

A chemist conducted a study on ammonia NH3(g) into hydrogen gas and nitrogen gas. This reaction is represented by the following

balanced equation:
2 NH3(g) à N2(g) + 3 H2(g)

The rate of consumption of ammonia gas was determined to be 0.0036 mol/(Ls). What is the rate of formation for hydrogen gas in the experiment?
Chemistry
1 answer:
Lelu [443]2 years ago
7 0

Answer:

.

Explanation:

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Answer:

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3 years ago
How many molecules are in 2.3 moles of h2o
Naddik [55]

Each mole of substance contains 6.02 x <span>1023</span> component parts, in this case water molecules.

If you have 2.3 moles of water you will have 2.3 x 6.02 x <span>1023</span> which is 1.3846 x <span>1024</span> molecules.

Each molecule contains 2 hydrogen atoms, so the total number of hydrogen atoms in 2.3 moles of water will be 2 x 1.3846 x <span>1024</span> = 2.7692 x <span>1024</span>.

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7 0
3 years ago
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How many moles are in 5g of Al?<br><br>A.0.185<br>B.18.5<br>C.5.4<br>D.135
Lapatulllka [165]
I got 134.91 but if you round it you’ll get 135
3 0
3 years ago
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A 5.76 liter sample of a gas at 22.00C mL 748 torr pressure was heated to a final volume of 17.28 liters, with the pressure rema
timama [110]

Answer:

The final temperature was 612 °C

Explanation:

Charles's law relates the volume and temperature of a certain amount of ideal gas, maintained at a constant pressure, using a constant of direct proportionality. In this law, Charles says that at constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. That is, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the ratio between volume and temperature will always have the same value:

\frac{V}{T}=k

When you want to study two different states, an initial and a final one of a gas and evaluate the change in volume as a function of temperature or vice versa, you can use the expression:

\frac{V1}{T1} =\frac{V2}{T2}

In this case:

  • V1= 5.76 L
  • T1= 22 °C= 295 °K (Being 0°C=273°K)
  • V2=17.28 L
  • T2=?

Replacing:

\frac{5.76 L}{295 K} =\frac{17.28 L}{T2}

Solving:

T2 =\frac{17.28 L}{\frac{5.76 L}{295 K}}=\frac{17.28 L*295 K}{5.76 L}

T2= 885 °K = 612 °C

<u><em>The final temperature was 612 °C</em></u>

5 0
3 years ago
Determine the density of a liquid from the following data: Mass of the
Tanya [424]

Answer:0.477 g/ml

Explanation:

Density=(40.14-33.79)/13.3 ml

Density=6.35/13.3

Density=0.477 g/ml

6 0
3 years ago
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