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1 year ago

How many atoms are in each of the following molecules:A. CO2B. N2C. CHCOOH

Chemistry

1 answer:

zlopas [31]1 year ago

4 0

<h2>Answer:</h2>

A) 3 atoms - 1 atom of Carbon and 2 atoms of oxygen.

B) 2 atoms of Nitrogen.

C) 6 atoms - 2 Carbon atoms, 2 Hydrogen atoms, and 2 Oxygen atoms.

<h2>Explanations:</h2>

A molecule is a group of atoms bonded together, representing the smallest fundamental unit of a chemical compound. Molecules are made up of atoms.

According to the following information, we are to find the number of atoms in the given molecules.

A) For carbon dioxide CO₂, this molecule is made of 3 atoms - 1 atom of Carbon and 2 atoms of oxygen.

B) For the compound N₂, this molecule is made up of 2 atoms of Nitrogen.

C) For the compound CHCOOH, this molecule consists of 6 atoms - 2 Carbon atoms, 2 Hydrogen atoms, and 2 Oxygen atoms.

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2 years ago

Exactly how much time must elapse before 16 grams of potassium-42decays, leaving 2 grams of the original isotope?(1) 8 × 12.4 ho

AleksandrR [38]

The answer is <span>(3) 3 × 12.4 hours
</span>
To calculate this, we will use two equations:
$(1/2) ^{n} =x$
$t_{1/2} = \frac{t}{n}$
where:
<span>n - number of half-lives
</span>x - remained amount of the sample, in decimals
<span> $t_{1/2}$ - half-life length
</span>t - total time elapsed.

First, we have to calculate x and n. x is <span>remained amount of the sample, so if at the beginning were 16 grams of potassium-42, and now it remained 2 grams, then x is:
2 grams : x % = 16 grams : 100 %
x = 2 grams </span>× 100 percent ÷ 16 grams
x = 12.5% = 0.125

Thus:
<span> $(1/2) ^{n} =x$
</span> $(0.5) ^{n} =0.125$
$n*log(0.5)=log(0.125)$
$n= \frac{log(0.5)}{log(0.125)}$
$n=3$

It is known that the half-life of potassium-42 is 12.36 ≈ 12.4 hours.
Thus:
<span> $t_{1/2} = 12.4$
</span><span> $t_{1/2} *n = t$
</span> $t= 12.4*3$

Therefore, it must elapse 3 × 12.4 hours <span>before 16 grams of potassium-42 decays, leaving 2 grams of the original isotope</span>

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3 years ago

Read 2 more answers