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3 years ago

Help meee pleaseeeee

Chemistry

1 answer:

tatiyna3 years ago

7 0

Answer:

see explanation

Explanation:

Using the periodic table, look at the top of each column => These are group numbers ... Typically (for american tables) the numbers are divided into A-Groups and B-Groups... For this post, you need to focus on the A-Groups, or 'Main Group Elements'... Now, the numbers also represent the number of valence (outer shell) electrons in the elements listed below that number. For example, under Group IA, all elements ( H, Li, Na, K, Rb, Cs & F) all have one (1) outer shell electron. All elements under IIA have two outer shell electrons, IIIA, 3 outer shell electrons and so on. The exception is Helium (He) which has only 2 outer shell electrons and is typically listed under Group VIIIA.

So ...

X· => H, Li & Na

X: => He(noble gas exception), Be & Mg

·X: => B & Al

:X: => C & Si

X(5 dots) => N

X(6 dots) => O

X(7 dots) => F & I

X(8 dots) => Ne

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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

Calculate the equilibrium constant K for the following reaction: H2(g) +

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Answer:

192.9

Explanation:

From the question,

Ke = [HCL]²/[H₂][CL₂].......................... Equation 1

Where Ke = Equilibrium constant.

Given: [HCL] = 0.0625 M, [H₂] = 0.0045 M, [CL₂] = 0.0045 M

Substitute these values into equation 1

Ke = (0.0625)²/(0.0045)(0.0045)

ke = (3.90625×10⁻³)/(2.025×10⁻⁵)

ke = 1.929×10²

ke = 192.9

Hence the equilibrium constant of the system = 192.9

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