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GrogVix [38]
3 years ago
9

The gases listed are at standard temperature and pressure: · 1 mol of CO2 gas · 1 mol of N2 gas · 2 mol of O2 gas Which statemen

t is correct? The volume of 1 mol of CO2 is greater than that of 2 mol of O2 . The volume of 1 mol of CO2 is greater than that of 1 mol of N2 . The volumes of 2 mol of O2 and 1 mol of N2 gases are the same. **The volumes of 1 mol of CO2 and 1 mol of N2 gases are the same.**
Chemistry
2 answers:
andrezito [222]3 years ago
8 0
The right answer for the question that is being asked and shown above is that: "The volumes of 1 mol of CO2 and 1 mol of N2 gases are the same." The <span>statement that is correct as far as the </span>gases listed are at standard temperature and pressure is concerned is that t<span>he volumes of 1 mol of CO2 and 1 mol of N2 gases are the same</span>
Elena L [17]3 years ago
6 0

Answer : The volume of 1 mole of CO_{2} and N_{2} gases are the same.

Solution: Given,

Moles of CO_{2} = 1 mole

Moles of N_{2} = 1 mole

Moles of O_{2} = 2 mole

Formula used for ideal gas is :

P V = n R T

According to the question, the gases are at standard temperature and pressure. So, the volume of gases only depends on the number of moles. This means that the higher the number of moles, higher will be the volume of gas.

The moles of O_{2} are more than the moles of CO_{2} and N_{2}. So, the volume of O_{2} will be more.

And the moles of CO_{2} and N_{2} are equal. Therefore, their volumes are also equal.

Therefore, the best option is the volume of 1 mole of CO_{2} and N_{2} gases are the same.

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Draw the structures of two stereoisomeric alkenes that would give 3-hexanol as the only major product of hydroboration
larisa [96]

Answer:

Cis- and trans-3-hexene are symmetric hydrocarons that give only one major product i.e 3-hexanol upon hydroboration.

Explanation:

During hydroboration of 3-hexene, borane (BH3) is added to the double bond of hexene,  that transfers the hydrogen atoms to that carbon which becomes is bonded to the boron. The process of hydroboration is created in two steps that leads to the formation of 3-hexanol and boric acid. (please see figure)

Now, the two stereoisomers, Cis- and trans-3-hexene both will give off the 3-hexanol upon hydroboration and the structure of these are illustrated in the figure.



4 0
3 years ago
Read 2 more answers
Correct way for calculating atomic mass
kolezko [41]

Answer:

mass number = protons + neutrons.

Explanation:

Together, the number of protons and the number of neutrons determine an element's mass number: mass number = protons + neutrons. If you want to calculate how many neutrons an atom has, you can simply subtract the number of protons, or atomic number, from the mass number.

5 0
2 years ago
Why can’t you ice skate on a lake when it is not frozen?
snow_lady [41]
Because you’ll fall in the water.
6 0
3 years ago
Iron is biologically important in the transport of oxygen by red blood cells from the lungs to the various organs of the body. I
Aneli [31]

Answer : The number of iron atoms present in each red blood cell are, 1.077\times 10^9

Explanation :

First we have to calculate the moles of iron.

\text{Moles of iron}=\frac{\text{Mass of iron}}{\text{Molar mass of iron}}=\frac{2.90g}{55.85g/mole}=0.0519moles

Now we have to calculate the number of iron atoms.

As, 1 mole of iron contains 6.022\times 10^{23} number of iron atoms

So, 0.0519 mole of iron contains 0.0519\times 6.022\times 10^{23}=3.125\times 10^{22} number of iron atoms

Now we have to calculate the number of iron atoms are present in each red blood cell.

Number of iron atoms are present in each red blood cell = \frac{\text{Number of iron atoms}}{\text{Total number of red blood cells}}

Number of iron atoms are present in each red blood cell = \frac{3.125\times 10^{22}}{2.90\times 10^{13}}

Number of iron atoms are present in each red blood cell = 1.077\times 10^9

Therefore, the number of iron atoms present in each red blood cell are, 1.077\times 10^9

6 0
3 years ago
What is the % yield when 140.0 grams of Ethylene gas (C2H4) reacts with excess chlorine to form 280.0 grams of 1,2-Dichloro Etha
KatRina [158]

Answer:

percent yield = 56.6 %

Explanation:

Data given:

mass of  Ethylene gas (C₂H₄) = 140 g

actual yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)= 280 g

percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂) = ?

Reaction Given:

                        C₂H₄ + Cl₂ -------> C₂H₄Cl₂

Solution:

First we have to find theoretical yield.

So,

Look at the reaction

                       C₂H₄ + Cl₂ -----—> C₂H₄Cl₂

                      1 mol                         1 mol

As 1 mole of C give 1 mole of CH₄

Convert moles to mass

molar mass of C₂H₄ = 2(12) + 4(1)

molar mass of C₂H₄ = 24 + 4

  • molar mass of C₂H₄ = 28 g/mol

molar mass of C₂H₄Cl₂ = 2(12) + 4(1) + 2(35.5)

molar mass of C₂H₄Cl₂ = 24 + 4 + 71

  • molar mass of C₂H₄Cl₂ = 99 g/mol

Now

                       C₂H₄     +       Cl₂    -----—>     C₂H₄Cl₂

                1 mol (28 g/mol)                        1 mol (99 g/mol)

                          28 g                                          99 g

28 grams of Ethylene gas (C₂H₄) produce 99 grams of C₂H₄Cl₂

So

if 28 g of C₂H₄ produce 99 g of C₂H₄Cl₂ so how many grams of C₂H₄Cl₂ will be produced by 140 g of C₂H₄.

Apply Unity Formula

                        28 g of C₂H₄ ≅ 99 g of C₂H₄Cl₂

                        140 g of C₂H₄≅ X of C₂H₄Cl₂

Do cross multiply

                      mass of C₂H₄Cl₂ = 99 g x 140 g / 28 g

                      mass of C₂H₄Cl₂ = 495 g

So the Theoretical yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)= 495 g

Now Find the percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂)

Formula Used

percent yield = actual yield /theoretical yield x 100 %

Put value in the above formula

                percent yield = 280g / 495 g x 100 %

                 percent yield = 56.6 %

percent yield of 1,2-Dichloro Ethane (C₂H₄Cl₂) = 56.6 %

8 0
4 years ago
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