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olga nikolaevna [1]
3 years ago
11

Which gases in the following list exist as separate atoms: hydrogen, helium, nitrogen, oxygen, neon.

Chemistry
2 answers:
iris [78.8K]3 years ago
6 0

helium and neon ( answer C)

lbvjy [14]3 years ago
6 0
The answer is C tell me if I was wrong
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How are nitrogen-14 and nitrogen-15 same and different?
Likurg_2 [28]

Answer:

Since different isotopes of an element have different numbers of neutrons (but always the same number of protons) they have different mass numbers. Nitrogen-14 and nitrogen-15 are both stable isotopes of nitrogen. However, the other 5 isotopes are all unstable.

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The prophet Isaac was chosen to lead the Israelites after the death of Abraham.
Softa [21]
False. . . . . . . .
8 0
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Calculate the missing variables in each experiment below using Avogadro’s law.
blagie [28]

Answer:

The answer to your question is: letter c

Explanation:

Data

V1 = 612 ml    n1 = 9.11 mol

V2 = 123 ml    n2 = ?

Formula

                               \frac{V1}{n1}  =  \frac{V2}{n2}

                                         n2 = \frac{n1V2}{V1}

                                         n2 = \frac{(9.11)((123)}{(612)}

                                                n2 = 1.83 mol                                                

5 0
3 years ago
You have 100 mL of a solution of benzoic acid in water; the amount of benzoic acid in the solution is estimated to be about 0.30
dimaraw [331]

Answer:

0.00370 g

Explanation:

From the given information:

To determine the amount of acid remaining using the formula:\dfrac{(final \ mass \ of \ solute)_{water}}{(initial \ mass \ of \ solute )_{water}} = (\dfrac{v_2}{v_1+v_2\times k_d})^n

where;

v_1 = volume of organic solvent = 20-mL

n = numbers of extractions = 4

v_2 = actual volume of water = 100-mL

k_d = distribution coefficient = 10

∴

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +20 \ ml \times 10})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{100 \ ml}{100 \ ml +200 \ ml})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = (\dfrac{1}{3})^4

\dfrac{(final \ mass \ of \ solute)_{water}}{0.30  \ g} = 0.012345

Thus, the final amount of acid left in the water = 0.012345 * 0.30

= 0.00370 g

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What is a variable and what do it do
frosja888 [35]
Varuable, In algebra, a symbol (usually a letter) standing in for an unknown numerical value in an equation :)
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