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2 years ago

Which of the following is an example of the law of multiple proportions?

1 answer:

3 0

The answer option which is an example of the law of multiple proportions is: B) Two different compounds formed from carbon and oxygen have the following mass ratios: 1.33 g O: 1 g C and 2.66 g O: 1 g C.

The Law of Multiple Proportions is also referred to as Dalton's Law and it states that when two chemical elements combine to form more than one chemical compound, the masses (weights) of one chemical element that combine with a fixed mass (weight) of the other chemical element will always be in a ratio of small whole numbers.

According to the Law of Multiple Proportions, when two chemical elements combine to form two different chemical compound, the masses (weights) of one chemical element that combine with 1 gram of the other chemical element can be expressed as a ratio of small whole numbers.

For example, carbon and oxygen react to form either carbon monoxide and carbon dioxide.

Carbon monoxide (CO):

12 grams of C = 16 grams of O.

1 gram of C = 1.33 gram of O.

Carbon dioxide ( $CO_2$ ):

12 gram of C = 32 grams of O.

1 gram of C = 2.66 gram of O.

Ratio of oxygen (O) = 16:32 = 1:2 (Law of Multiple Proportions).

Your question is lacking the necessary answer options, so I will be adding them here:

A. A sample of chlorine is found to contain three times as much Cl-35 as Cl-37.

B. Two different compounds formed from carbon and oxygen have the following mass ratios: 1.33 g O: 1 g C and 2.66 g O: 1 g C.

C. Two different samples of table salt are found to have the same ratio of sodium to chlorine.

D. The atomic mass of bromine is found to be 79.90 amu.

E. Nitrogen dioxide always has a mass ratio of 2.28 g O: 1 g N.

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In other words the acceleration can be described as

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At the same time since Newton's second law we know that:

$F_w = ma$

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Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:

$dF_W = mda$

$dF_W = m(-2\frac{GM}{r^3}dr)$

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But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:

$dF_W = -2(600)(\frac{1.6*10^3}{6.37*10^6})$

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Therefore there is a weight loss of 0.3N every kilometer.

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Vanyuwa [196]

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18 N + F₃ = -12 N

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