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nika2105 [10]
3 years ago
7

Write the balanced decomposition reaction of the hypochlorite ion into the chlorate and chloride ions using the smallest integer

coefficients. (in this reaction, sodium ions are spectator ions.)
Chemistry
1 answer:
Alja [10]3 years ago
3 0

This is a redox reaction:

2OCl- --> ClO2- + Cl-

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To determine the concentration of one solution which is specifically basic or acidic solution through taking advantage on its points of equivalence, titration analysis is done.

Let us determine the reaction for the titration below:

2NaOH +2H2SO4 = Na2SO4 +2H2O

 

So,

0.0665 mol NaOH (2 mol H2SO4/ 2mol NaOH) / .025 L solution 

= 2.62 M H2SO4


The answer is the fourth option: 

<span>2.62 M</span>
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How many significant figures are in the measurement, 0.005890 g?
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4 significant figures. Use a rule called the Atlantic-Pacific rule where if the period is absent (Atlantic), you would start counting the numbers from right to left. If the period is present (Pacific), then start counting the numbers from left to right. Since the period is present, start from 5 and there are 4 digits including 5.

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Help in my exam plzz now​
swat32

Answer:

yes your answer is correct

I hope it helps mate

I will always help you understanding your assingments have a great day

#Captainpower:)

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3 years ago
In each case, calculate the appropriate ratio to show that the information given is consistent with the law of multiple proporti
Alchen [17]

Answer:

(a) 3:2; (b) 2:1

Explanation:

The Law of Multiple Proportions states that when two elements A and B combine to form two or more compounds, the masses of B that combine with a given mass of A are in the ratios of small whole numbers.

That is, if one compound has a ratio r₁ and the other has a ratio r₂, the ratio of the ratios r is in small whole numbers.

(a) Ammonia and hydrazine.

In ammonia, the mass ratio of H:N is r₁ = 0.2158/1

In hydrazine, the mass ratio of H:N is r₂ = 0.1439/1

The ratio of the ratios is:

r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 0.2158}{0.1439} = \dfrac{1.500 }{1} = \dfrac{2.999}{2} \approx \mathbf{\dfrac{3}{2}}\\\\\text{The relative amounts of H in the two compounds are in the ratio }\boxed{\mathbf{\dfrac{3}{2}}}

(b) Nitrogen oxides

In nitrogen monoxide, the mass ratio of O:N is r₁ = 1.142/1

In dinitrogen monoxide, the mass ratio of O:N is r₂ = 0.571/1

The ratio of the ratios is:

r = \dfrac{r_{1}}{r_{2}} = \dfrac{ 1.142}{0.571} = \dfrac{2.000 }{1} \approx \mathbf{\dfrac{2}{1}}\\\\\text{The relative amounts of O in the two compounds are in the ratio }\boxed{\mathbf{\dfrac{2}{1}}}

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3 years ago
Pls help me with this question
Drupady [299]
The answer is c hopefully I helped you
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