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3 years ago

The chemical equation for the formation of ammonia is unbalanced.

1 answer:

7 0

Balance Chemical equation is as follow,

 3 H₂ (g) + N₂ (g) → 2 NH₃ (g)

According to balanced equation, 3 Molecules (3 moles) of Hydrogen reacts with 1 Molecule of N₂ to produce 2 moles (2 Molecules) of NH₃.

Result:
2 Molecules of Ammonia are produced by reacting 3 molecules of Hydrogen and 1 molecule of Nitrogen.

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How much pressure would 0.8 moles of a gas at 370K exert if it occupied 17.3L of space

dezoksy [38]

Answer:

1.40 atm is the pressure for the gas

Explanation:

An easy problem to solve with the Ideal Gases Law:

P . V = n . R .T

T° = 370K

V = 17.3L

n = 0.8 mol

Let's replace data → P . 17.3L = 0.8mol . 0.082L.atm/mol.K . 370K

P = (0.8mol . 0.082L.atm/mol.K . 370K) / 17.3L = 1.40 atm

7 0

3 years ago

Read 2 more answers

PLEASE HELP i’ll mark brainliest ( reporting fake answers)

kherson [118]

Delta S reaction= Delta S products- Delta S reactants

don't forget to mulitiply by coefficients

also

here is a really slow way to do it

you know the moles of gas increased

so Delta S is positive

so its B or D

then just do the units digit to see which one match up

6 0

3 years ago

Theory that the lithosphere is divided into tectonic plates that slowly move on

vitfil [10]

Pute because I’m smart and I know the right answer

5 0

3 years ago

Determine the mole fractions and partial pressures of CO2, CH4, and He in a sample of gas that contains 1.20 moles of CO2, 1.79

BARSIC [14]

Answer : The mole fraction and partial pressure of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179, 0.554 and 1.54, 1.03 and 3.20 atm respectively.

Explanation : Given,

Moles of $CH_4$ = 1.79 mole

Moles of $CO_2$ = 1.20 mole

Moles of $He$ = 3.71 mole

Now we have to calculate the mole fraction of $CH_4,CO_2$ and $He$ gases.

$\text{Mole fraction of }CH_4=\frac{\text{Moles of }CH_4}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CH_4=\frac{1.79}{1.79+1.20+3.71}=0.267$

and,

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }CO_2=\frac{1.20}{1.79+1.20+3.71}=0.179$

and,

$\text{Mole fraction of }He=\frac{\text{Moles of }He}{\text{Moles of }CH_4+\text{Moles of }CO_2+\text{Moles of }He}$

$\text{Mole fraction of }He=\frac{3.71}{1.79+1.20+3.71}=0.554$

Thus, the mole fraction of $CH_4,CO_2$ and $He$ gases are, 0.267, 0.179 and 0.554 respectively.

Now we have to calculate the partial pressure of $CH_4,CO_2$ and $He$ gases.

According to the Raoult's law,

$p_i=X_i\times p_T$

where,

$p_i$ = partial pressure of gas

$p_T$ = total pressure of gas = 5.78 atm

$X_i$ = mole fraction of gas

$p_{CH_4}=X_{CH_4}\times p_T$

$p_{CH_4}=0.267\times 5.78atm=1.54atm$

and,

$p_{CO_2}=X_{CO_2}\times p_T$

$p_{CO_2}=0.179\times 5.78atm=1.03atm$

and,

$p_{He}=X_{He}\times p_T$

$p_{He}=0.554\times 5.78atm=3.20atm$

Thus, the partial pressure of $CH_4,CO_2$ and $He$ gases are, 1.54, 1.03 and 3.20 atm respectively.

4 0

3 years ago

What three countries are considered the ‘big three’ when it comes to space programs and are the only countries to successfully l

Colt1911 [192]

Answer:

C. United States, Russia, and China

Explanation:

Only three nations (Russia, U.S., China) have launched their own crewed spacecraft, with the Soviets/Russians and the American programs providing rides to other nations' astronauts. Twenty-seven "first flights" occurred on Soviet or Russian flights while the United States carried fourteen.

8 0

2 years ago