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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?
Chemistry
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
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