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KATRIN_1 [288]
3 years ago
6

How much 0.075 M NaCl solution can be made by diluting 450 mL of 9.0 M NaCl?

Chemistry
1 answer:
mafiozo [28]3 years ago
4 0

Answer:

54 L

Explanation:

Given data

  • Initial concentration (C₁): 9.0 M
  • Initial volume (V₁): 450 mL
  • Final concentration (C₂): 0.075 M
  • Final volume (V₂): to be calculated

We have a concentrated NaCl solution and we will add water to it to obtain a diluted NaCl solution. We can find the volume of the final solution using the dilution rule.

C_1 \times V_1 = C_2 \times V_2\\V_2 = \frac{C_1 \times V_1}{C_2} = \frac{9.0M \times 450mL}{0.075M} = 5.4 \times 10^{4} mL = 54 L

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3 years ago
How many grams of FeCo3 will be produced from 57.2g FeCl2
Evgesh-ka [11]

Answer:

             287.30 g of FeCO₃

Solution:

The Balance Chemical Equation is as follow,

                           FeCl₂ + Na₂CO₃    →    FeCO₃ + 2 NaCl

Step 1: Calculate Mass of FeCl₂ as,

                            Molarity  =  Moles ÷ Volume

Solving for Moles,

                            Moles  =  Molarity × Volume

Putting Values,

                            Moles  =  2 mol.L⁻¹ × 1.24 L

                           Moles  =  2.48 mol

Also,

                            Moles  =  Mass ÷ M.Mass

Solving for Mass,

                            Mass  =  Moles × M.Mass

Putting Values,

                            Mass  =  2.48 mol × 126.75 g.mol⁻¹

                            Mass =  314.34 g of FeCl₂

Step 2: Calculate Mass of FeCO₃ formed as,

According to equation,

          126.75 g (1 mole) FeCl₂ produces  =  115.85 g (1 mole) FeCO₃

So,

               314.34 g of FeCl₂ will produce  =  X g of FeCO₃

Solving for X,

                     X =  (314.34 g × 115.85 g) ÷ 126.75 g

                     X =  287.30 g of FeCO₃

<h2>brainlyest pleas</h2>
4 0
2 years ago
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