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zmey [24]
2 years ago
8

How is this type of question solved?

Chemistry
1 answer:
SIZIF [17.4K]2 years ago
5 0

The percentage purity of the sodium hydroxide will be 37.5%.

option A is correct.

<h3>What is mass percentage?</h3>

Mass percentage is the amount of solute present in per hundred gram of a solution is called Mass percentage of solution.

Mass percentage = mass of component / mass of solution × 100

mass of component = 1.20g

mass of solution = molar mass of NaOH + molar mass of HCl

                            = 40 + 36.5

                             = 76.5

substituting the value in the equation,

mass percentage = 1.20 / 76.5

                             = 37.5%

Therefore, percentage purity of the sodium hydroxide will be 37.5%.

option A is correct.

learn more about mass percentage, here:

brainly.com/question/14848802

#SPJ1

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Determine the mass of hydrogen contained in 9.06 x 1024 H2O molecules.
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Answer:

15.2 g H2

Explanation:

2H2O -> 2H2 + O2

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Hydrogen is manufactured on an industrial scale by this sequence of reactions: Write an equation that gives the overall equilibr
RideAnS [48]

The question is incomplete. The complete question is :

Hydrogen is manufactured on an industrial scale by this sequence of reactions:

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

The net reaction is  :

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

Write an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K_1 and K_2. If you need to include any physical constants, be sure you use their standard symbols, which you'll find in the ALEKS Calculator.

Solution :

$CH_3(g) + H_2O(g) \rightleftharpoons CO(g) + 3H_2(g    ) \ \ \ \ \ \ \ \ \ \ K_1$

$K_1 = \frac{[CO][H_2]^3}{[CH_4][H_2O]}$     ...............(1)

$CO(g) + H_2O(g) \rightleftharpoons CO_2(g) + H_2(g) \ \ \ \ \ \ \ \ \ \ \ \  K_2$

$K_2 = \frac{[CO_2][H_2]}{[CO][H_2O]}$  ...................(2)

$CH_4(g) + 2H_2O(g) \rightleftharpoons CO_2(g) + 4H_2(g) \ \ \ \ \ \ \ \ \ K$

$K=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$

On multiplication of equation (1) and (2), we get

$K_1 \times K_2=\frac{[CO][H_2]^3}{[CH_4][H_2O]} \times \frac{[CO_2][H_2]}{[CO][H_2O]}$

$K_1K_2=\frac{[CO_2][H_2]^4}{[CH_4][H_2O]^2}$  .................(4)

Comparing equation (3) and equation (4), we get

$K=K_1K_2$

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2 years ago
Balance this equation. If a coefficient of "1" is required, choose "blank" for that box.
Kitty [74]

Answer: 2 C2H4 + 6 O2 => 4 CO2 + 4 H2O

Explanation:The coefficient are as follows: 2: 6: 4: 4

Each atom on the reactant and product side are equal.

Reactant Product

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H 2x4 = 8 4x2 = 8

O 6x2 = 12 (4x2) + 4 = 12

7 0
3 years ago
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