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postnew [5]
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.​

Chemistry
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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Dvinal [7]

Answer :  The percent abundance of Li isotope-1 and Li isotope-2 is, 6.94 % and 93.1 % respectively.

Explanation :

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i   .....(1)

Let the fractional abundance of Li isotope-1 be 'x' and the fractional abundance of Li isotope-2 will be '100-x'

For Li isotope-1 :

Mass of Li isotope-1 = 6.01512 amu

Fractional abundance of Li isotope-1 = x

For Li isotope-2 :

Mass of Li isotope-2 = 7.01600 amu

Fractional abundance of Li isotope-2 = 100-x

Average atomic mass of Li = 6.941 amu

Putting values in equation 1, we get:

6.941=[(6.01512\times x)+(7.01600\times (100-x))]

By solving the term 'x', we get:

x=694.048

Percent abundance of Li isotope-1 = \frac{694.048}{100}=6.94\%

Percent abundance of Li isotope-2 = 100 - x = 100-6.94 = 93.1 %

6 0
3 years ago
Find the grams in 1.26 x 10^-4 mole of HC2 H3O2
ivolga24 [154]
Molar mass :

HC₂H₃O₂ = 1 + 12*2 + 1 * 3 + 16 * 2 = 60 g/mol

1 mole <span>HC₂H₃O₂ -------------- 60 g
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mass =  1.26x10-⁴ * 60

mass = 0.00756 g of <span>HC₂H₃O₂</span>

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7 0
3 years ago
There is 35 mg of sodium in a can of Coke. You determine it to be 28 mg. What is your percent error?
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Answer:

The answer is

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Explanation:

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actual volume = 35 mg

error = 35 - 28 = 7

The percentage error is

P(\%) =  \frac{7}{35}  \times 100 \\  =  \frac{1}{5}  \times 100 \\  =  20

We have the final answer as

<h3>20 %</h3>

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