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2 years ago

12

Amol has divided some substances into two groups as shown in the table below. He wants to further separate all the substances in

Group 1 again into exactly two groups. Which of the following criteria will allow him to do so?
A. liquids and solids

B. mixtures and pure substances

C. naturally occurring and man-made substances.

D. substances completely soluble and insoluble in water.

2 answers:

Nonamiya [84]2 years ago

6 0

The criteria that will allow Amol separate group 1 substances into two groups are mixtures and pure substances (option B).

<h3>What are mixtures and pure substances?</h3>

Mixtures are substances that are made up of two or more substances that can be easily separated.

Pure substances, on the other hand, are elements and compounds that cannot be separated by chemical means.

In group 1 of the above table, the substances can be classified as follows:

Soil - mixture
Tomato ketchup - mixture
Smoke - compound
Milk - mixture
Hair shampoo - mixture

Therefore, the criteria that will allow Amol separate group 1 substances into two groups are mixtures and pure substances.

Learn more about mixture at: brainly.com/question/24898889

#SPJ1

Degger [83]2 years ago

5 0

Answer:

Explanation:

b is the right answer

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Methylpropane (complete and condensed structural formulas)

Tcecarenko [31]

Answer:

Here's what I get

Explanation:

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The complete structural formula is shown below (There is a C atom at each intersection).

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A gaseous mixture composed of 20% CH4, 30% C2H4, 35% C2H2, and 15% C2H20. What is the average molecular weight of the mixture? a

Semenov [28]

<u>Answer:</u> The correct answer is Option d.

<u>Explanation:</u>

We are given:

Mass percentage of $CH_4$ = 20 %

So, mole fraction of $CH_4$ = 0.2

Mass percentage of $C_2H_4$ = 30 %

So, mole fraction of $C_2H_4$ = 0.3

Mass percentage of $C_2H_2$ = 35 %

So, mole fraction of $C_2H_2$ = 0.35

Mass percentage of $C_2H_2O$ = 15 %

So, mole fraction of $C_2H_2O$ = 0.15

We know that:

Molar mass of $CH_4$ = 16 g/mol

Molar mass of $C_2H_4$ = 28 g/mol

Molar mass of $C_2H_2$ = 26 g/mol

Molar mass of $C_2H_2O$ = 48 g/mol

To calculate the average molecular mass of the mixture, we use the equation:

$\text{Average molecular weight of mixture}=\frac{_{i=1}^n\sum{\chi_im_i}}{n_i}$

where,

$\chi_i$ = mole fractions of i-th species

$m_i$ = molar masses of i-th species

$n_i$ = number of observations

Putting values in above equation:

$\text{Average molecular weight}=\frac{(\chi_{CH_4}\times M_{CH_4})+(\chi_{C_2H_4}\times M_{C_2H_4})+(\chi_{C_2H_2}\times M_{C_2H_2})+(\chi_{C_2H_2O}\times M_{C_2H_2O})}{4}$

$\text{Average molecular weight of mixture}=\frac{(0.20\times 16)+(0.30\times 28)+(0.35\times 26)+(0.15\times 42)}{4}\\\\\text{Average molecular weight of mixture}=6.75$

Hence, the correct answer is Option d.

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Answer:

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