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Kay [80]
4 years ago
7

Compounds A and BB are colorless gases obtained by combining sulfur with oxygen. Compound A results from combining 6.00 gg of su

lfur with 5.99 gg of oxygen, and compound BB results from combining 8.60 gg of sulfur with 12.88 gg of oxygen. Show that the mass ratios in the two compounds are simple multiples of each other.
Chemistry
1 answer:
Lelechka [254]4 years ago
5 0

Answer:

Mass ratio of sulfur and oxygen in compounds A and B is 3:2 which confirms that the mass ratios in the two compounds are simple multiples of each other

Explanation:

This question seeks to establish/confirm the law of multiple proportions which  posits that elements combine to form different substances which are whole number multiples of each other. Best example of this plays out in the formation of several oxides of the same element. Looking at the ratio in which the elements combine in each of the oxides, we can assume that these ratios are simple whole number multiples of each other.

Now back to the question.

In substance A, we have 6 g of sulfur combining with 5.99 g of oxygen

Now, lest us calculate the ratio of the mass of sulfur to that of oxygen = 6g/5.99g = 1

Now let us calculate the mass ratio of sulfur to oxygen in the second compound = 8.6/12.88 = 0.668

Now the ratios in both compounds are 1 to 0.668. 0.668 to fraction is approximately 1/1.5.

So therefore, the ratio we are having would be 1:1/1.5 or 1:0.668

This is same as 1/(1÷1.5) which is 1.5/1 or simply 3/2

This gives a ratio of approximately 1.5 to 1 or 3 to 2

The ratio 3 to 2 indicates that the mass ratios in both com pounds are simple multiples of each other

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Explanation:

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For each of the esters provided, identify the alcohol and the carboxylic acid that reacted.
Veronika [31]

Answer:

52. The alcohol USED => methanol, CH3OH

The carboxylic acid USED => propanoic acid, CH3CH2COOH.

53. The alcohol USED => Ethanol, CH2CH3OH

The carboxylic acid USED => Formic acid, HCOOH.

Explanation:

52. To obtain Methyl propanoate, CH3CH2COOCH3, we simply react propanoic, CH3CH2COOH and methanol, CH3OH together as shown below:

CH3CH2COOH + CH3OH —> CH3CH2COOCH3 + H2O

The alcohol used: methanol, CH3OH

The carboxylic acid used: propanoic acid, CH3CH2COOH.

53. To obtain Ethyl methanoate, HCOOCH2CH3, we simply react

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HCOOH + CH3CH2OH —> HCOOCH2CH3 + H2O

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