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meriva
3 years ago
7

Which statement below BEST describes the difference between mixtures and pure substances? Group of answer choices Mixtures are m

ade of a single kind of matter but a pure substance is two or more substances that are chemically bonded. Pure substances can be physically separated but mixtures cannot be physically separated. Pure substances are not created through a chemical reaction and mixtures are the result of a chemical reaction. Mixtures can be physically separated but pure substances cannot be physically separated.
Chemistry
1 answer:
lianna [129]3 years ago
6 0

Answer:  <u>Mixtures can be physically separated but pure substances cannot be physically separated.</u>

Explanation: Mixtures can be physically separated by using methods that use differences in physical properties to separate the components of the mixture, such as evaporation, distillation, filtration and chromatography. A pure substance, any compound or element, can't be separated into different atoms by physical methods. Pure substances are compounds and elements (made up of the same atom or same molecule respectively), while mixtures are an assortment of different substances put together. Hope this helps ^-^.

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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.

3 0
3 years ago
I need help please by answering that question
jenyasd209 [6]
If i can recall it is D
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Answer:

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Which determine the charge of an atom?
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Answer:

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