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2 years ago

7

If you trace back the history of a carbon atom in your little finger through all of cosmic history, where did this atom most lik

ely originate?.

1 answer:

Kamila [148]2 years ago

8 0

Answer:

It was fused from 3 helium nuclei in the core of a red giant star long before the Sun existed

Explanation:

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Magnesium reacts with carbon dioxide to form magnesium oxide and pure solid carbon.

guajiro [1.7K]

Answer:

2Mg + 2CO₂ → 2MgO + C

Option a.

Explanation:

The chemical reactions are balanced in order to verify compliance with the Law of conservation of mass. This law states that in any chemical reaction, the mass of the reagents must be equal to the mass of the products

7 0

3 years ago

What new information can you add to your definition of physical change ?

Mamont248 [21]

Its reversible, soluble, <span>mass, density, color, boiling point, temperature, and volume. </span>

7 0

3 years ago

Which atom would it be most difficult to remove an electron from

Norma-Jean [14]

I would be difficult to remove an electron from a Noble or Inert Gas (also known as the group 8 or 0 elements). This is because they all have filled outermost shells and as such the outermost shell would be held tightly to the nucleus and as such make it difficult to remove. Examples Helium, Neon, Argon, Xenon, Krypton and Radon

6 0

3 years ago

A mixture of helium and methane gases, at a total pressure of 821 mm Hg, contains 0.723 grams of helium and 3.43 grams of methan

Nookie1986 [14]

<u>Answer:</u>

<u>For 1:</u> The partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

<u>For 2:</u> The mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

<u>Explanation:</u>

<u>For 1:</u>

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

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For helium:</u>

Given mass of helium = 0.723 g

Molar mass of helium = 4 g/mol

Putting values in equation 1, we get:

$\text{Moles of helium}=\frac{0.723g}{4g/mol}=0.181mol$

<u>For methane gas:</u>

Given mass of methane gas = 3.43 g

Molar mass of methane gas = 16 g/mol

Putting values in equation 1, we get:

$\text{Moles of methane gas}=\frac{3.43g}{16g/mol}=0.214mol$

To calculate the mole fraction , we use the equation:

$\chi_A=\frac{n_A}{n_A+n_B}$ .......(2)

To calculate the partial pressure of gas, we use the equation given by Raoult's law, which is:

$p_{A}=p_T\times \chi_{A}$ ......(3)

<u>For Helium gas:</u>

We are given:

$n_{He}=0.181mol\\n_{CH_4}=0.214mol$

Putting values in equation 2, we get:

$\chi_{He}=\frac{0.181}{0.181+0.214}=0.458$

Calculating the partial pressure by using equation 3, we get:

$p_T=821mmHg\\\\\chi_{He}=0.458$

Putting values in equation 3, we get:

$p_{He}=0.458\times 821mmHg=376mmHg$

<u>For Methane gas:</u>

We are given:

$n_{He}=0.181mol\\n_{CH_4}=0.214mol$

Putting values in equation 2, we get:

$\chi_{CH_4}=\frac{0.214}{0.181+0.214}=0.542$

Calculating the partial pressure by using equation 3, we get:

$p_T=821mmHg\\\\\chi_{CH_4}=0.542$

Putting values in equation 3, we get:

$p_{CH_4}=0.542\times 821mmHg=445mmHg$

Hence, the partial pressure of helium is 376 mmHg and that of methane gas is 445 mmHg

<u>For 2:</u>

We are given:

Partial pressure of nitrogen gas = 363 mmHg

Partial pressure of carbon dioxide gas = 564 mmHg

Total pressure = (363 + 564) mmHg = 927 mmHg

Calculating the mole fraction of the gases by using equation 3:

<u>For nitrogen gas:</u>

$363=\chi_{N_2}\times 927\\\\\chi_{N_2}=\frac{363}{927}=0.392$

<u>For carbon dioxide gas:</u>

$564=\chi_{CO_2}\times 927\\\\\chi_{CO_2}=\frac{564}{927}=0.608$

Hence, the mole fraction of nitrogen gas is 0.392 and that of carbon dioxide gas is 0.608

6 0

3 years ago

Problem Page Liquid hexane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 70. g of

shusha [124]

Answer:

The maximum mass of carbon dioxide that could be produced by the chemical reaction is 70.6gCO_{2}

Explanation:

1. Write down the balanced chemical reaction:

$2C_{6}H_{14}_{(l)}+19O_{2}_{(g)}=12CO_{2}_{(g)}+14H_{2}O_{(g)}$

2. Find the limiting reagent:

- First calculate the number of moles of hexane and oxygen with the mass given by the problem.

For the hexane:

$70.0gC_{6}H_{14}*\frac{1molC_{6}H_{14}}{86.2gC_{6}H_{14}}=0.81molesC_{6}H_{14}$

For the oxygen:

$81.3gO_{2}*\frac{1molO_{2}}{32.0gO_{2}}=2.54molesO_{2}$

- Then divide the number of moles between the stoichiometric coefficient:

For the hexane:

$\frac{0.81}{2}=0.41$

For the oxygen:

$\frac{2.54}{19}=0.13$

- As the fraction for the oxygen is the smallest, the oxygen is the limiting reagent.

3. Calculate the maximum mass of carbon dioxide that could be produced by the chemical reaction:

The calculations must be done with the limiting reagent, that is the oxygen.

$81.3gO_{2}*\frac{1molO_{2}}{32gO_{2}}*\frac{12molesCO_{2}}{19molesO_{2}}*\frac{44.0gCO_{2}}{1molCO_{2}}=70.6gCO_{2}$

7 0

3 years ago