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ruslelena [56]
2 years ago
15

1.Calculate the mass of compound required to prepare 100.00 mL of a 1.40x10-2M CuCl2.Show your work.

Chemistry
1 answer:
Verdich [7]2 years ago
5 0

The concentration of the diluted solution is 0.0625 M.

<h3>What is concentration?</h3>

The term concentration refers to the amount of substance present in solution.

1) To find the mass of the compound;

mass/134 g/mol = 1.40x10-2M × 100/1000

mass = 1.40x10-2M × 100/1000 × 134 g/mol

mass = 0.19 g

2) To find the molarity of the solution;

2.50 g/40 = 100/1000 × M

M = 2.50 g/40 × 1000/100

M = 0.625 M

3) Using;

C1V1 = C2V2

0.625 M × 10.00mL = C2 × 100.00 mL

C2 = 0.625 M × 10.00mL/ 100.00 mL

C2 = 0.0625 M

Learn more about concentration: brainly.com/question/3045247

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PLEASE HELP!!!
Romashka-Z-Leto [24]

Answer:

43.76 g = 0.04376 Kg

Explanation:

The equation of the reaction is given as;

H2 + Cl2 --> 2HCl

From the equation;

1 mol of Cl2 produces 2 mol of HCl

Converting to mass;

Mass = Number of mol * Molar mass

Cl2;

Mass = 1 mol * 70.906 g/mol = 70.906 g

HCl;

Mass = 2 mol * 36.458 g/mol = 72.916 g

This means;

70.906 g of Cl2 produces 72.916 g of HCl

x g of Cl2 would produce 45g of HCl

70.906 = 72.916

x = 45

x = 45 * 70.906 / 72.916

x = 43.76g

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Explanation:

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6 0
3 years ago
A 0.500-g sample of KCl is added to 50.0g of water in a calprimeter (Figure 5.12) If the temperature decreases by 1.05C. what is
Sauron [17]

Answer : The reaction is endothermic.

Explanation :

Formula used :

Q=m\times c\times \Delta T

where,

\Delta T = change in temperature = 1.05^oC

Q = heat involved in the dissolution of KCl = ?

m = mass = 0.500 + 50.0 = 50.5 g

c = specific heat of resulting solution = 4.18J/g^oC

Now put all the given value in the above formula, we get:

Q=50.5g\times 4.18J/g^oC\times 1.05^oC

Q=+221.64J

The heat involved in the dissolution of KCl is positive that means as the change in temperature decreases then the reaction is endothermic and as the change in temperature increases then the reaction is exothermic.

Hence, the reaction is endothermic.

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