What would happen if an atom of fluorine (F) came into contact with an atom of neon (Ne)?

Chemistry

1 answer:

juin [17]3 years ago

7 0

A. nothing neon is a noble gas, meaning that it dosnt react with anything

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Below is a bar graph on candy allowance compared to the child's happiness.

rjkz [21]

Answer:

There is no bar graph attached to this question, however, the question can be answered based on the information given in the question.

The answer is A) average level of happiness

Explanation:

In an experiment, the dependent variable is the variable which is measured by the experimenter. It is the variable that responds to changes made to another variable called independent variable.

In the case of this question, it can be determined, even without the bar graph, that the experiment entails how candy allowance affects a child's happiness. Hence, the candy allowance is changed to influence or cause a response in the child's happiness, which is then measured. Therefore, the AVERAGE LEVEL OF HAPPINESS is the dependent variable.

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3 years ago

Write balanced equations for each of the processes described below. (Use the lowest possible coefficients. Omit states-of-matter

Aleks [24]

Answer:

Explanation:

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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:

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