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lbvjy [14]
3 years ago
15

Explain why mixtures are examples of a physical change. Why aren’t they a chemical change? What separates a physical change from

a chemical change?
Help Pleaseee!!
Chemistry
1 answer:
s2008m [1.1K]3 years ago
6 0

Answer:

This is a physical change because it's not permanent and no chemical reaction occurs. Mixtures - Mixing together materials where one is not soluble in the other is a physical change. Note the properties of a mixture may be different from its components.

Explanation:

The difference between a physical reaction and a chemical reaction is composition. In a chemical reaction, there is a change in the composition of the substances in question; in a physical change there is a difference in the appearance, smell, or simple display of a sample of matter without a change in composition.

In a physical change, the molecules are rearranged while their actual composition remains same. In a chemical change, the molecular composition of a substance completely changes and a new substance is formed. Some example of physical change are freezing of water, melting of wax, boiling of water....

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A chemist places 2.5316 g of Na 2SO 4 in a 100 mL volumetric flask and adds water to the mark. She then pipets 15 mL of the resu
Lera25 [3.4K]

Answer:

The concentration of the most dilute solution is 0.016M.

Explanation:

First, a solution is prepared and then it undergoes two subsequent dilutions. Let us calculate initial concentration:

[Na_{2}SO_{4}]=\frac{moles(Na_{2}SO_{4})}{liters(solution)} =\frac{mass((Na_{2}SO_{4}))}{molarmass(moles(Na_{2}SO_{4}) \times 0.100L)} =\frac{2.5316g}{142g/mol\times 0.100L } =0.178M

<u>First dilution</u>

We can use the dilution rule:

C₁ x V₁ = C₂ x V₂

where

Ci are the concentrations

Vi are the volumes

1 and 2 refer to initial and final state, respectively.

In the first dilution,

C₁ = 0.178 M

V₁ = 15 mL

C₂ = unknown

V₂ = 50 mL

Then,

C_{2}=\frac{C_{1} \times V_{1} }{V_{2}} =\frac{0.178M \times 15mL}{50mL} =0.053M

<u>Second dilution</u>

C₁ = 0.053 M

V₁ = 15 mL

C₂ = unknown

V₂ = 50 mL

Then,

C_{2}=\frac{C_{1} \times V_{1} }{V_{2}} =\frac{0.053M \times 15mL}{50mL} =0.016M

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