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antoniya [11.8K]

3 years ago

14

After your done with the second step,you will be left with a clear colorless liquid. What should be done to the liquid to finish

separation?

1 answer:

mafiozo [28]3 years ago

5 0

In order to retrieve the soluble salt from the solution, crystallization must be carried out.<span />

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How many formula units are in 5.33 moles of cucl2

Blizzard [7]

Answer:

32.1 × 10²³ formula units of CuCl₂

Explanation:

Given data:

Number of moles of CuCl₂ = 5.33 mol

Number of formula units of CuCl₂ = ?

Solution:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

one mole of any substance contain 6.022 × 10²³ formula units thus,

5.33 moles of CuCl₂ = 5.33 ×6.022 × 10²³ formula units

32.1 × 10²³ formula units of CuCl₂

3 0

3 years ago

A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o

statuscvo [17]

<u>Answer:</u> The value of $K_{eq}$ is 0.044

<u>Explanation:</u>

We are given:

Initial moles of methane = $4.10\times 10^{-2}mol=0.0410moles$

Initial moles of carbon tetrachloride = $6.51\times 10^{-2}mol=0.0651moles$

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 methane = $\frac{0.0410}{1.00}=0.0410M$

Concentration of carbon tetrachloride = $\frac{0.0651}{1.00}=0.0651M$

The given chemical equation follows:

$CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)$

<u>Initial:</u> 0.0410 0.0651

<u>At eqllm:</u> 0.0410-x 0.0651-x 2x

We are given:

Equilibrium concentration of carbon tetrachloride = $6.02\times 10^{-2}M=0.0602M$

Evaluating the value of 'x', we get:

$\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M$

Now, equilibrium concentration of methane = $0.0410-x=[0.0410-0.0049]=0.0361M$

Equilibrium concentration of $CH_2Cl_2=2x=[2\times 0.0049]=0.0098M$

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

$K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044$

Hence, the value of $K_{eq}$ is 0.044

8 0

3 years ago

*WILL GIVE BRAINLIEST TO CORRECT ANSWER *<br> Question in picture

sineoko [7]

The Answer for this question is : A

1 & 2

5 0

2 years ago

Read 2 more answers

Brianliest for whoever gets this right with step by step method

Sedaia [141]

The Mr is the mass numbers of each element added up so…. Fe = 56, O=16, H=1 … now add these up with the number of each element -> there’s 1 Fe, and 3 Os and 3 Hs as they are in brackets with a 3 outside-> (56+16+16+16+1+1+1=107) … your answer is 107

8 0

3 years ago

Which must be the same when comparing 1 mol of oxygen gas, O2, with 1 mol of carbon monoxide gas, CO?

Ber [7]

Answer: C. The Number of Molecules

Explanation: I just took the test, it's correct.

6 0

3 years ago