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3 years ago

Kc= [3.110^-2]^2. (mol/L^2) divide by [8.510^-1] [3.1*10^-3]^3 (mol/L)^4

Chemistry

1 answer:

jok3333 [9.3K]3 years ago

5 0

Answer:

Kc = 12.58

Explanation:

Kc = [0.229]^2*[0.687]^6/[0.221]^4*[0.5685]^3

Kc = (0.052441)(0.10513)/(0.002385)(0.18373)

Kc = 0.0005513/0.000438

Kc = 12.58

Hope that helps!!

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Which of the following is an anion?

satela [25.4K]

A . O2- because it is a negatively charged ion

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3 years ago

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 $4.84\times 10^{-5}$

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Answer:

Rosalind Elsie Franklin (25 July 1920 – 16 April 1958)was a British biophysicist and X-ray crystallographer who made critical contributions to the understanding of the fine molecular structures of DNA, RNA, viruses, coal and graphite. The DNA work achieved the most fame because DNA (deoxyribonucleic acid) plays essential roles in cell metabolism and genetics, and the discovery of its structure helped scientists understand how genetic information is passed from parents to children.

rosalindfranklin

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3 0

2 years ago

If 0. 3 moles of H2O were produced, how many moles of carbon dioxide would also have been produced?

grigory [225]

In a chemical reaction, 0. 3 moles of H2O result in the production of 0.2 moles of CO2. Chemical reactions are the means through which one group of chemical compounds are changed into another.

Chemical reactions are typically defined as changes that only affect the positions of electrons in the formation and breakage of chemical bonds between atoms, with no change to the nuclei (i.e., no change to the elements present). These types of changes are often included in the term chemical reactions.

Number of moles Definition We utilize this enormous quantity to measure atoms. Additionally, it equals the 6.022* 10 23 atoms that make up 12 grammes of carbon-12, or atoms.

C2H6 + 7/2O2 = 2CO2 + 3H2O,

where moles(CO2)=(2*0,3)/3=0.2 mol,

and n(CO2)=(2*0,3)/3=n(H2O);

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1 year ago

An electron in the hydrogen atom makes a transition from an energy state of principal quantum number ni to the n = 2 state. If t

topjm [15]

Answer:

$\boxed{3}$

Explanation:

The Rydberg equation gives the wavelength λ for the transitions:

$\dfrac{1}{\lambda} = R \left ( \dfrac{1}{n_{i}^{2}} - \dfrac{1}{n_{f}^{2}} \right )$

where

R= the Rydberg constant (1.0974 ×10⁷ m⁻¹) and

$\text{$n_{i}$ and $n_{f}$ are the numbers of the energy levels}$

Data:

$n_{f} = 2$

λ = 657 nm

Calculation:

$\begin{array}{rcl}\dfrac{1}{657 \times 10^{-9}} & = & 1.0974 \times 10^{7}\left ( \dfrac{1}{2^{2}} - \dfrac{1}{n_{f}^{2}} \right )\\\\1.522 \times 10^{6} &= &1.0974\times10^{7}\left(\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \right )\\\\0.1387 & = &\dfrac{1}{4} - \dfrac{1}{n_{f}^{2}} \\\\-0.1113 & = & -\dfrac{1}{n_{f}^{2}} \\\\n_{f}^{2} & = & \dfrac{1}{0.1113}\\\\n_{f}^{2} & = & 8.98\\n_{f} & = & 2.997 \approx \mathbf{3}\\\end{array}\\\text{The value of $n_{i}$ is }\boxed{\mathbf{3}}$

7 0

3 years ago

Kc= [3.1*10^-2]^2. (mol/L^2) divide by [8.5*10^-1] [3.1*10^-3]^3 (mol/L)^4​

Kc= [3.110^-2]^2. (mol/L^2) divide by [8.510^-1] [3.1*10^-3]^3 (mol/L)^4