What is the name of the functional group that is attached to this hydrocarbon?

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

What is the answers?

ra1l [238]

the numbers are going to be small so like a power but its at the bottom

NH3, H2O2, NHO2

5 0

3 years ago

What is [h3o ] in a solution of 0.075 m hno2 and 0.030 m nano2?

Tanya [424]

Answer is: concentration of hydrogenium ions is 9,54·10⁻⁵ M.
c(HNO₂) = 0,075 M.
c(NaNO₂) = 0,035 M.
Ka(HNO₂) = 4,5·10⁻⁵.
This is buffer solution, so use Henderson–Hasselbalch equation:
pH = pKa + log(c(NaNO₂) ÷ c(HNO₂)).
pH = -log(4,5·10⁻⁵) + log(0,035 M ÷ 0,075 M).
pH = 4,35 - 0,33.
pH = 4,02.
[H₃O⁺] = 10∧(-4,02).
[H₃O⁺] = 0,0000954 M = 9,54·10⁻⁵ M.

8 0

3 years ago

A galaxy contains stars, gas, and dust. This matter must be held together or the galaxy will break apart. What is responsible fo

Marrrta [24]

Gravity is responsible !

5 0

3 years ago

How many milliliters of 5.0 M NaOH are needed to exactly neutralize 40. milliliters of 2.0 M HCl?

zubka84 [21]

Answer:

16mL

Explanation:

Using the following formula;

CaVa = CbVb

Where;

Where

Ca = concentration/molarity of acid (M)

Va = volume of acid (mL)

Cb = concentration/molarity of base (M)

Vb = volume of base (mL)

According to the information provided in this question;

Ca (HCl) = 2M

Cb (NaOH) = 5M

Va (HCl) = 40mL

Vb (NaOH) = ?

Using CaVa = CbVb

Vb = CaVa/Cb

Vb = 2 × 40/5

Vb = 80/5

Vb = 16mL

5 0

2 years ago