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2 years ago

Can someone please help me.

Chemistry

1 answer:

siniylev [52]2 years ago

5 0

Answer:

Sure, with what?

Ps- Just to increase word count

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Force acts on it.

BartSMP [9]

Answer:D.

Explanation:

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4 years ago

Breathing air that contains 4.0% by volume CO2 over time causes rapid breathing, throbbing headache, and nausea, among other sym

7nadin3 [17]

Answer:

1 mole of any ideal gas occupies the same volume as one mole of any other ideal gas under the same conditions of temperature and pressure.So 1 L CO2 has same number of moles as 1 L O2 and 1 L N2 etc.This means that, assuming all the gases in air are ideal gases, if CO2 is 4.0 % by volume then it is also 4.0 % by moles, because a volume of each gas has the same number of moles.

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3 years ago

In the reaction between Li and O, there is a transfer of electrons making an ionic bond. In the bond, lithium would be a (n) bec

galben [10]

Answer:

C) cation, loses, anion, gains

Explanation:

Lithium is a metal from Group 1, so it has 1 valence electron. Thus, it loses 1 electron to complete its octet and form the cation Li⁺.

Oxygen is a nonmetal from Group 16, so it has 6 valence electrons. Thus, it gains 2 electrons to complete its octet and form the anion O²⁻.

In the reaction between Li and O, there is a transfer of electrons making an ionic bond. In the bond, lithium would be a cation because it loses an electron, and oxygen would be an anion because it gains 2 electrons.

6 0

4 years ago

The total mass of the atmosphere is about 5.00 x 1018 kg. How many moles each of air, O2, and CO2 are present in the atmosphere?

n200080 [17]

Answer: The moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of atmosphere = $5.00\times 10^{18}kg=5.00\times 10^{21}g$

Average molar mass of atmosphere = 28.96 g/mol

Putting values in above equation, we get:

$\text{Moles of atmosphere}=\frac{5.00\times 10^{21}g}{28.96g/mol}=1.73\times 10^{20}mol$

We know that:

Percent of oxygen in air = 21 %

Percent of carbon dioxide in air = 0.0415 %

Moles of oxygen in air = $\frac{21}{100}\times 1.73\times 10^{20}=3.63\times 10^{19}mol$

Moles of carbon dioxide in air = $\frac{0.0415}{100}\times 1.73\times 10^{20}=7.18\times 10^{16}mol$

Hence, the moles of oxygen and carbon dioxide in air is $3.63\times 10^{19}mol$ and $7.18\times 10^{16}mol$ respectively

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3 years ago

Matthew and Bradley got into an argument. Matthew was convinced that the above picture was RNA, and Bradley was positive that it

sashaice [31]

the right answer is .C )Matthew was right because RNA is single-stranded while DNA is double-stranded.

8 0

3 years ago

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