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IRISSAK [1]
3 years ago
13

For the reaction

Chemistry
1 answer:
azamat3 years ago
7 0

Answer:

Mass = 5.56 g

Explanation:

Given data:

Mass of Cl₂ = 4.45 g

Mass of NaCl produced = ?

Solution:

Chemical equation:

2Cl₂ + 4NaOH     →   3NaCl + NaClO₂ + 2H₂O

Number of moles of Cl₂:

Number of moles = mass/molar mass

Number of moles = 4.45 g/ 71 g/mol

Number of moles = 0.063 mol

Now we will compare the moles of Cl₂ with NaCl.

                  Cl₂         :         NaCl

                    2          :          3

                 0.063      :        3/2×0.063 =0.095 mol

Mass of NaCl:

Mass = number of moles × molar mass

Mass = 0.095 mol × 58.5 g/mol

Mass = 5.56 g

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Answer:

CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓

Explanation:

We identify the reactants:

CuBr₂ and Pb(CH₃COO)₂

The products will be: Cu(CH₃COO)₂ and PbBr₂

You may know these information:

Salts from acetate are soluble.

Bromide can make solid salts with these cations: Ag⁺, Pb²⁺, Hg₂²⁺, Cu⁺

PbBr₂ is formed, so this will be our precipitate

The equation is:

CuBr₂(aq) + Pb(CH₃COO)₂(aq) → Cu(CH₃COO)₂(aq) + PbBr₂ (s)↓

8 0
3 years ago
What is the correct name for: C3S5?
scZoUnD [109]

Answer:

Trithioxodithiole

Explanation:

5 0
3 years ago
An element is to a __ as an organ is to a ___
Vera_Pavlovna [14]

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An element is to a compound as an organ is to tissue

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5 0
3 years ago
A stock solution of sodium sulfate NaSO4 has a concentrate of 1.00 m. The volume of this solution is 50 ml. What volume of 0.25
mylen [45]
<h3>Answer:</h3>

200 mL

<h3>Explanation:</h3>

Concept tested: Dilution formula

We are given;

  • Concentration of stock solution as 1.00 M
  • Volume of the stock solution as 50 mL
  • Molarity of the dilute solution as 0.25 M

We are required to calculate the volume of diluted solution;

  • The stock solution is the original solution before dilution while diluted solution is the solution after dilution.
  • Using the dilution formula we can determine the volume of diluted solution;

M1V1 = M2V2

Rearranging the formula;

V2 = M1V1 ÷ M2

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     = 200 mL

Therefore, a volume of 200mL of 0.25 M solution could be made from the stock solution.

5 0
3 years ago
a) What substances are present in an aqueous buffer composed of HC2H3O2 and C2H3O2 - ?b) What happens when LiOH is added to a bu
Alex17521 [72]

Answer:

a) HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻

b) HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

Explanation:

a) In a HC₂H₃O₂/C₂H₃O₂⁻ buffer system, the following reactions take place:

HC₂H₃O₂ + H₂O ⇄ C₂H₃O₂⁻ + H₃O⁺

C₂H₃O₂⁻ + H₂O ⇄ HC₂H₃O₂ + OH⁻

Thus, the species present are: HC₂H₃O₂, C₂H₃O₂⁻, H₃O⁺, H₂O, OH⁻.

b) When LiOH is added to the buffer system, it is partially neutralized according to the following equation.

HC₂H₃O₂ + LiOH ⇄ H₂O + LiC₂H₃O₂

c) When HBr is added to the buffer system, it is partially neutralized according to the following equation.

C₂H₃O₂⁻ + HBr ⇄ HC₂H₃O₂ + Br⁻

3 0
3 years ago
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