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Natalka [10]
3 years ago
10

One brand of laundry bleach is an aqueous solution containing 4.00% sodium hypochlorite (naocl) by mass. you may want to referen

ce (pages 586 - 592) section 13.5 while completing this problem. part a what is the molarity of this solution? (assume a density of 1.02 g/ml.)
Chemistry
2 answers:
Zanzabum3 years ago
5 0

Answer:  B. 0.599

Explanation: If you are using usatestprep 8^)

Vlad1618 [11]3 years ago
3 0
Answer is: <span>molarity of this solution is 0,55 mol/l.
</span>If we use 100 ml of solution:
V(solution) = 100 ml ÷ 1000 ml/l = 0,1 l.
ω(NaClO) = 4,00% = 0,04.
d(solution) = 1,02 g/ml.
m(solution) = V(solution) · d(solution).
m(solution) = 100 ml · 1,02 g/ml.
m(solution) = 102 g.
m(NaClO) = ω(NaClO) · m(solution).
m(NaClO) = 0,04 · 102 g = 4,08 g.
n(NaClO) = m(NaClO) ÷ M(NaClO).
n(NaClO) = 4,08 g ÷ 74,44 g/mol = 0,055 mol.
c(NaClO) = 0,055 mol ÷ 0,1 l = 0,55 mol/l.
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you find that 7.36g of a compound has decomposed to give 6.93g of oxygen. the only other element in the compound is hydrogen. if
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3 years ago
You take three compounds consisting of two elements and decompose them. To determine the relative masses of X, Y, and Z, you col
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Answer:

a) LAW OF MULTIPLE PROPORTIONS

b) 0.095g, 0.71g, 0.285g respectively

c) X2Y, YZ15, X6Y

d) hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

Explanation:

a) The assumptions made in solving this questions is the application of the LAW OF MULTIPLE PROPORTIONS. The Law of multiple proportions states that if two elements A and B combine together to form more than one compound, then the several masses of A which chemically combine with a fixed mass of B is in a simple ratio.

for example, copper forms two oxides ; copper(I) oxide (CuO) and copper(ii) oxide(Cu2O), it is possible for the two samples of the oxides to be reduced to Cu by reacting with Hydrogen gas. as such, certain masses of oxygen combine separately with a fixed mass of Cu. then the ratios of Cu are then determined.

b) To calculate the relative masses, we take note of the three compounds given, they all have some amount of Y in them, hence we can use Y  as our relative mass, this implies that the relative mass of Y = 1g

mass of X = 0.4g

mass of Y = 4.2g

amount of X in 1g of Y = 0.4 x 1 /4.2

= 0.095g

for compound 2;

mass of Y = 1.4g

mass of Z = 1.0g

amount of Z in 1g of Y =1.0 x 1 /1.4

= 0.71g

for compound 3;

mass of X = 2.0g

mass of Y = 7.0g

amount of X in 1g of Y = 1 x 2/7

= 0.285g

c) Applying the law of multiple proportions; since elements X and Z combine with a fixed mass of Y, they must bear a simple ratio;

compound 1/compound 3 = 0.095/0.285

= 1/3

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= 2/15

compound 2/ compound 3 = 0.71/0.285

= 5/2

formular for compound 1 = X2Y

formula for compound 2 = YZ15

formular for compound 3 = X6Y

d) from the formular X2Y, we can get the amount of each product in XY using the ratios

%of compound XY in X = mass of compound X / total Mass

= 0.2/4.4 = 4.5%

as such in a 21g of compound XY, %of compound Y = 1 - %of compound X = 95.5%

hence mass of compound X = 21 x 0.045 = 0.95g

mass of compound Y = 21 x 0.955 = 20.05g

5 0
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