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Ahat [919]
3 years ago
6

What is the molarity of a solution containing 45 grams of CaCl2 in 1.5 L of water?

Chemistry
1 answer:
s2008m [1.1K]3 years ago
7 0

Answer:

Data:

Mass of solute: 45 g

m.mass of solute: 110 g/mol

volume of water: 1.5 L

Explanation:

percent w/v: 45/1500 *100

vol of solution: 3cm3=0.003dm3

M=g of solute/m.mass of solute*dm3 of sol

M= 45/110*0.003

M=14.85

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Answer: Mass of CO_2  produced in this reaction was 6.56 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

CaCO_3(s)+2HCl(aq)\rightarrow H_2O(l)+CO_2(g)+CaCl_2(aq)

Mass or reactants =  Mass of CaCO_3+ mass of HCl = 16.00 + 64.80 = 80.80 g

Mass of products  = mass of aqueous solution + mass of CO_2 + = 74.24 + x g

Mass or reactants = Mass of products

80.80 g = 74.24 + x g

x = 6.56 g

Thus mass of CO_2  produced in this reaction was 6.56 grams

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3 years ago
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What do tissues combine to form?
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A pure substance made up of molecules of the same kind of atoms is called...
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All substances that are  made up of the same kind of mole ules are called pure substances.

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The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s
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Answer : The equilibrium concentration of H_3O^+ in the solution is, 2.1\times 10^{-3}M

Explanation :

The dissociation of acid reaction is:

                       C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-

Initial conc.        c                                 0                0

At eqm.             c-x                                 x                x

Given:

c = 7.0\times 10^{-2}M

K_a=6.3\times 10^{-5}

The expression of dissociation constant of acid is:

K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}

K_a=\frac{(x)\times (x)}{(c-x)}

Now put all the given values in this expression, we get:

6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}

x=2.1\times 10^{-3}M

Thus, the equilibrium concentration of H_3O^+ in the solution is, 2.1\times 10^{-3}M

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3 years ago
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