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2 years ago

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Examples of pure and impure substances

2 answers:

erastova [34]2 years ago

8 0

Pure- table salt
Impure- vegetable oil

Sladkaya [172]2 years ago

8 0

Examples of pure substances are water, diamond, table salt

Examples of impure substances are muddy water, air, sugar in water

<u>Explanation</u>:

A pure substance is one that is made up of only one element or one chemical compound. It has no impurities. They can be elements or compounds.
It has definite properties and composition.
Examples are water, diamond, table salt, etc.
An impure substance is one that is made up of two or more elements or compounds that are not chemically bonded. It is a mixture.
It has no definite composition or properties.
Examples are muddy water, sugar in water, air, etc.

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Consider the second-order reaction:

kirza4 [7]

Answer:

Initial concentration of HI is 5 mol/L.

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

Explanation:

$2HI(g)\rightarrow H_2(g)+I_2(g)
$

Rate Law: $k[HI]^2
$

Rate constant of the reaction = k = $6.4\times 10^{-9} L/mol s$

Order of the reaction = 2

Initial rate of reaction = $R=1.6\times 10^{-7} Mol/L s$

Initial concentration of HI = $[A_o]$

$1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2$

$[A_o]=5 mol/L$

Final concentration of HI after t = [A]

t = $4.53\times 10^{10} s$

Integrated rate law for second order kinetics is given by:

$\frac{1}{[A]}=kt+\frac{1}{[A_o]}$

$\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}$

$[A]=0.00345 mol/L$

The concentration of HI after $4.53\times 10^{10} s$ is 0.00345 mol/L.

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How many kinds of atoms are in a pure substance?

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2 years ago

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Part 1: What is the final volume in milliliters when 0.730 L of a 44.8 % (m/v) solution is diluted to 23.3 % (m/v)?

Andre45 [30]

part 1 : the final volume : 1.404 L

part 2 : the initial concentration : 4.06 M

<h3>Further explanation </h3>

Dilution is the process of adding a solvent to get a more dilute solution.

The moles(n) before and after dilution are the same.

Can be formulated :

M₁V₁=M₂V₂

M₁ = Molarity of the solution before dilution

V₁ = volume of the solution before dilution

M₂ = Molarity of the solution after dilution

V₂ = Molarity volume of the solution after dilution

part 1 :

M₁=44.8%

V₁=0.73 L

M₂=23.3%

$\tt V_2=\dfrac{M_1.V_1}{M_2}\\\\V_2=\dfrac{44.8\times 0.73}{23.3}\\\\V_2=1.404~L$

part 2 :

V₁=739 ml=0.739 L

V₂=1.5 L

M₂=2

$\tt M_1=\dfrac{M_2.V_2}{V_1}\\\\M_1=\dfrac{2\times 1.5}{0.739}\\\\M_1=4.06$

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2 years ago