SOMEONE PLEASE HELP HURRY

The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different react

murzikaleks [220]

Answer: The value of $K_{eq}$ is $4.84\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0280 moles

Initial moles of oxygen gas = 0.0120 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0280}{1.00}=0.0280M$

Concentration of oxygen gas = $\frac{0.0120}{1.00}=0.0120M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0280 0.0120

At eqllm: 0.0280-4x 0.0120-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $3.00\times 10^{-3}M=0.003$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.003\\\\\Rightarrow x=0.0015M$

Now, equilibrium concentration of ammonia = $0.0280-4x=[0.0280-(4\times 0.0015)]=0.022M$

Equilibrium concentration of oxygen gas = $0.0120-3x=[0.0120-(3\times 0.0015)]=0.0075M$

Equilibrium concentration of water = $6x=(6\times 0.0015)]=0.009M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[H_2O]^6\times [N_2]^2}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.009)^6\times (0.003)^2}{(0.022)^4\times (0.0075)^3}\\\\K_{eq}=4.84\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.84\times 10^{-5}$

4 0

3 years ago

Strong versus weak acids what makes a strong acid strong pogil answer key

Over [174]

An acid is deemed strong if it can readily or easy "donate" a proton (H+) to the other ions in the solutions. Also, to donate or lose the proton or H+, the acid must dissociate (split into ions) in the solution. The more it can readily dissociate, the stronger the acid is.

8 0

3 years ago

Read 2 more answers

If molar mass of M(OH)3 = 78 8. mass of M

Cerrena [4.2K]

Answer:

From molar mass=total RAM of each individual element

78.8=(16+1)×3+M

78.8-51=M

27.8g/mol=M

5 0

2 years ago

How many grams of CO are needed to react with an excess of Fe 2 O 3 to produce 209.7 g Fe?

gavmur [86]

Answer:

Mass = 157.5 g

Explanation:

Given data:

Mass of CO needed = ?

Mass of Fe formed = 209.7 g

Solution:

Chemical equation:

3CO + F₂O₃ → 2Fe + 3CO₂

Number of moles of Fe:

Number of moles = mass/ molar mass

Number of moles = 209.7 g/ 55.85 g/mol

Number of moles = 3.75 mol

Now we will compare the moles of iron and carbon monoxide.

Fe : CO

2 : 3

3.75 ; 3/2×3.75 = 5.625 mol

Mass of CO:

Mass = number of moles × molar mass

Mass = 5.625 mol × 28 g/mol

Mass = 157.5 g

4 0

2 years ago

Describe what occurs when a gas is collected by water displacement, including why a table of water vapor pressure values is need

Radda [10]

When a gas bubbles through water, small droplets of water are usually picked up along for the ride and are mixed in with the gas above the water inside the eudiometer tube. The water vapor takes up room, but isn't the important gas that you need to measure. The table of water vapor is needed to subtract the unwanted water vapor from the collection of gases.

5 0

2 years ago