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NemiM [27]
2 years ago
9

Suppose you need 0.12L of a 0.13M solution of NiCl2, but all you have is a solution that is a solution that is 0.23M. What volum

e of the solution do you need to dilute
Chemistry
1 answer:
topjm [15]2 years ago
3 0

- 0.05 L needs to be added to the original 0.12 L solution in order to dilute it from 0.13 M to 0.23 M.

The dilution problem uses the equation :

M_aV_a= M_bV_b

The initial molarity (concentration) M_a = 0.13 M

The initial volume V_a = 0.12 L

The desired molarity (concentration) M_b = 0.23 M

The volume of the desired solution V_b = ( 0.12 + x L )

Substituting values in above equation;

(0.13 M ) (0.12 L) = (0.23 M ) (0.12 L + x L)

0.0156 M L = 0.0276 M L + 0.23 x M L

- 0.012 M L = 0.23 x M L

x = - 0.05

Therefore, - 0.05 L needs to be added to the original 0.12 L solution in order to dilute it from 0.13 M to 0.23 M.

Learn more about molarity here:

brainly.com/question/26873446

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Explanation:

The electron configuration of an atomic species (neutral or ionic) allows us to understand the shape and energy of its electrons. Many general rules are taken into consideration when assigning the "location" of the electron to its prospective energy state, however these assignments are arbitrary and it is always uncertain as to which electron is being described. Knowing the electron configuration of a species gives us a better understanding of its bonding ability, magnetism and other chemical properties.

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3 years ago
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What is the molar mass of water in the hydrate? FeSO4 • 7H2O
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M=7M(H₂O)

M=7*18.015 g/mol = 126.105 g/mol
5 0
4 years ago
Determine the oxidation number of Cl in each of the following species.Cl2O7AlCl4-Ba(ClO2)2CIF4+
DIA [1.3K]

These are four questons and four answers:

Answers:

  • 1)  7⁺
  • 2) 1⁻
  • 3) 3⁺
  • 4) 5⁺

Explanation:

<u><em>Question 1) </em></u><u><em>Cl₂O₇:</em></u>

a) Net charge of the compound: 0

b) Rule: oxygen works with oxidation state +2, except with peroxides.

d) Rule: balance of charges: ∑ of the charges = net charge

Call X the oxidation number of Cl:

  • 2×X + 7 (-2) = 0
  • 2X - 14 = 0
  • 2X = +14
  • X = +14 /2 = + 7

<em>Conclusion: the oxidation number of Cl in Cl₂O₇ is 7⁺.</em>

<u><em>Question 2) </em></u><u><em>AlCl₄⁻</em></u>

a) Net charge of the ion: - 1

b) Rule: common oxidation number of Al in compounds: +3

c) Rule: balance of charges: ∑ charges = net charge = - 1

  • 1 (+3) + 4X = - 1
  • +3 + 4X = - 1
  • 4X = - 1 - 3
  • 4X = - 4
  • X = - 1

<em>Conclusion: the oxidation number of Cl in AlCl₄⁻ is 1 ⁻.</em>

<em><u>Question 3)</u></em><em><u> Ba(ClO₂)₂</u></em>

a) Net charge of the compound: 0

b) Rule: common oxidation number of BA in compounds: +2

c) Rule: common oxidation number of O in compounds (except in peroxides): -2

d) Rule: balance of charges: ∑ charges = net charge = 0

  • +2 + 2X + 4 (-2) = 0
  • 2X +2 - 8 = 0
  • 2X - 6 = 0
  • 2X = +6
  • X = + 3

<em>Conclusion: the oxidation number of Cl in Ba(ClO₂)₂  is 3⁺.</em>

<u><em>Question 4)</em></u><u><em> CIF₄⁺</em></u>

a) Net charge of the ion: + 1

b) Rule: common oxidation number of F : - 1 (it is the most electronegative)

c) Rule: balance of charges: ∑ charges = net charge = + 1

  • X + 4(-1) = +1
  • X - 4 = +1
  • X = +1 + 4
  • X = + 5

<em>Conclusion: the oxidation number of Cl in ClF₄⁺ is 5⁺.</em>

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The correct answer is option B. Dirty water is a mixture of solid particles and liquid. It is both a mixture and pure substance.

The dirty water sample has both gravel and liquid water in it. After filtration the gravel is removed so the water sample looks clearer than before filtration. Liquid water is a pure substance because it is a compound that is made up of elements hydrogen and oxygen. Now the gravel is only physically combined with the liquid water, thus giving the water sample properties of a mixture. And like any mixture, gravel is physically separated through filtration from the liquid water.

Thus the water sample of the chemists is both a mixture and pure substance.




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