How many atoms are in three moles of carbon?

a) 1.8 x 1023 atoms

b) 1.8 x 1024 atoms

c) 2.0 23 atoms

d) 18.0 atoms

Skip to content# How many atoms are in three moles of carbon? a) 1.8 x 1023 atoms b) 1.8 x 1024 atoms c) 2.0 23 atoms

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How many atoms are in three moles of carbon?

a) 1.8 x 1023 atoms

b) 1.8 x 1024 atoms

c) 2.0 23 atoms

d) 18.0 atoms

7.87 x 10²³ atoms

Explanation:

The number of LiCl units is related to moles using Avogadro's constant (6.022 × 10²³ mol⁻¹):

(0.654 mol)(6.022 × 10²³ mol⁻¹) = 3.93 x 10²³ LiCl

One LiCl unit is composed of two atoms, one Li atom and one Cl atom, so the amount is multiplied by 2:

(3.93 x 10²³)(2) = 7.87 x 10²³ atoms

Your answer should be b

1) B 2) both a and c

Answer : The correct option is, (a) [tex]4.52\times 10^{24}[/tex]

Explanation :

As we know that,

1 mole of substance contains [tex]6.022\times 10^{23}[/tex] number of atoms.

As we are given that:

The moles of [tex]CO_2[/tex] = 2.50 moles

Now we have to calculate the number of atoms present in [tex]CO_2[/tex] molecule.

In [tex]CO_2[/tex], there are 3 atoms in [tex]CO_2[/tex] molecule that means 1 atom of carbon and 2 atoms of oxygen.

As, 1 moles of [tex]CO_2[/tex] contains [tex]3\times 6.022\times 10^{23}[/tex] number of atoms.

So, 2.50 moles of [tex]CO_2[/tex] contains [tex]2.50\times 3\times 6.022\times 10^{23}=4.52\times 10^{24}[/tex] number of atoms.

Therefore, the number of atoms in 2.50 moles of [tex]CO_2[/tex] are [tex]4.52\times 10^{24}[/tex]

2.5 mol CO2 ( 2 mol O / 1 mol CO2 ) ( 6.022 x 10^23 atom / mole ) = 3.011 x 10^24 atoms of oxygen

Avogadro's number represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole. this could be converted.

7) the answer is D. You need to know that oxygen gas is O₂ which means that A and B can't be right. Then you need to see which one is balanced and since D is and C is not, the correct answer has to be D.

8) You need to know that 1 mole of anything contains 6.02×10²³ particles. therefore you multiply 6.02×10²³ by 3 to get 1.806×10²⁴ atoms.

9) You need to multiply the number of moles of carbon dioxide by the molar mass of carbon dioxide (44g/mol). 2.1mol×44g/mol=92.4g. Therefore the answer is 92.4g of CO₂.

I hope this helps. Let me know if anything is unclear.

B) 1.8 x 10^24 atoms

1.80 * 10^24 atoms is the answer your looking for.

1. How many moles are in 3.4 x 10²⁴ molecules of HCl?

if there are 6.022 × 10²³ molecules in 1 mole of HCl

then there are 3.4 × 10²⁴ molecules in X moles of HCl

X = (3.4 x 10²⁴ × 1) / 6.022 × 10²³ = 5.6 moles of HCl

2. How many grams are in 2.59 x 10²³ formula units of Al₂O₃?

if there are 6.022 × 10²³ units in 1 mole of Al₂O₃

then there are 2.59 × 10²³ units in X moles of Al₂O₃

X = (2.59 x 10²³ × 1) / 6.022 × 10²³ = 0.43 moles of Al₂O₃

number of moles = mass / molecular weight

mass = number of moles × molecular weight

mass of Al₂O₃ = 0.43 × 102 = 43.86 g

3. How many grams are in 9.05 x 1023 atoms of silicon?

if there are 6.022 × 10²³ atoms in 1 mole of silicon

then there are 9.05 × 10²³ atoms in X moles of silicon

X = (9.05 x 10²³ × 1) / 6.022 × 10²³ = 1.5 moles of silicon

mass = number of moles × molecular weight

mass of silicon = 1.5 × 28 = 42 g

4. If you had 7.00 moles of PtO₂, how many grams would you have?

mass = number of moles × molecular weight

mass of PtO₂ = 7 × 227 = 1589 g

5.How many moles are in 29.2 L of oxygen gas at STP?

At standard temperature and pressure (STP) we may use the following formula:

number of moles = volume (L) / 22.4 (L / mol)

number of moles of oxygen = 29.2 / 22.4 = 1.3

6. What is the volume (liters) of 75.8 g of N₂ at STP?

number of moles = mass / molecular weight

number of moles of N₂ = 75.8 / 28 = 2.7 moles

At standard temperature and pressure (STP) we may use the following formula:

number of moles = volume (L) / 22.4 (L / mol)

volume = number of moles × 22.4

volume of N₂ = 2.7 × 22.4 = 60.48 L

1. Answer is: there are 1.41·10²³ molecules of oxygen.

1) calculate amount of substance for oxygen gas:

m(O₂) = 7.5 g; mass of oxygen.

n(O₂) = m(O₂) ÷ M(O₂).

n(O₂) = 7.5 g ÷ 32 g/mol.

n(O₂) = 0.234 mol.

2) calculate number of molecules:

N(O₂) = n(O₂) · Na.

N(O₂) = 0.234 mol · 6.022·10²³ 1/mol.

N(O₂) = 1.41·10²³.

Na - Avogadro constant.

2. Answer is: the percent yield for the reaction is 61.77%.

Balanced chemical reaction: 2HgO(s) → 2Hg(l) + O₂(g).

m(HgO) = 4.37 g.

n(HgO) = n(HgO) ÷ M(HgO).

n(HgO) = 4.37 g ÷ 216.6 g/mol.

n(HgO) = 0.02 mol.

From chemical reaction: n(HgO) : n(Hg) = 2 : 2 (1 :1).

n(Hg) = n(HgO) = 0.02 mol; amount of substance.

m(Hg) = n(Hg) ·M(Hg).

m(Hg) = 0.02 mol · 200.6 g/mol.

m(Hg) = 4.047 g.

yield = 2.5 g ÷ 4.047 g · 100%.

yield = 61.77%.

3. Answer is: 68.16 grams of the excess reactant (oxygen) remain.

Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).

m(Fe) = 27.3 g.

n(Fe) = m(Fe) ÷ M(Fe).

n(Fe) = 27.3 g ÷ 55.85 g/mol.

n(Fe) = 0.489 mol.

m(O₂) = 79.9 g.

n(O₂) = 79.9 g ÷ 32 g/mol.

n(O₂) = 2.497 mol; amount of substance.

From chemical reaction: n(Fe) . n(O₂) = 4 : 3.

0.489 mol : n(O₂) = 4 : 3.

n(O₂) = 3 · 0.489 mol ÷ 4.

n(O₂) = 0.367 mol.

Δn(O₂) = 2.497 mol - 0.367 mol.

Δn(O₂) = 2.13 mol.

m(O₂) = 2.13 mol · 32 g/mol.

m(O₂) = 68.16 g.

4. Answer is: there are 0.603 moles of ammonia.

m(NH₃) = 10.25 g; mass of ammonia.

M(NH₃) = Ar(N) + 3Ar(H) · g/mol.

M(NH₃) = 14 + 3·1 · g/mol.

M(NH₃) = 17 g/mol; molar mass of ammonia.

n(NH₃) = m(NH₃) ÷ M(NH₃).

n(NH₃) = 10.25 g ÷ 17 g/mol.

n(NH₃) = 0.603 mol; amount of substance (ammonia).

5. Answer is: the empirical formula mass of P₂O₅ is 141.89.

Empirical formula gives the proportions of the elements present in a compound.

Atomic mass of phosphorus is 30.97 g/mol.

Atomic mass of oxygen is 15.99 g/mol.

In phosphorus (V) oxide there are two atoms of phosphorus and five atoms of oxygen:

EFM(P₂O₅) = 2·30.97 g/mol + 5·15.99 g/mol = 141.89 g/mol.

6. Answer is: there are 1.108·10²⁴ molecules of water.

n(H₂O) = 1.84 mol; amount of substance (water).

N(H₂O) = n(H₂O) · Na.

N(H₂O) = 1.84 mol · 6.022·10²³ 1/mol.

N(H₂O) = 11.08·10²³.

N(H₂O) = 1.108·10²⁴.

Na - Avogadro constant (number of particles (ions, atoms or molecules), that are contained in one mole of substance).

7. Answer is: iron (Fe) is the limiting reactant.

Balanced chemical reaction: 4Fe(s) + 3O₂(g)→ 2Fe₂O₃(g).

m(Fe) = 27.3 g.

n(Fe) = m(Fe) ÷ M(Fe).

n(Fe) = 27.3 g ÷ 55.85 g/mol.

n(Fe) = 0.489 mol.

m(O₂) = 45.8 g.

n(O₂) = 45.8 g ÷ 32 g/mol.

n(O₂) = 1.431 mol; amount of substance.

From chemical reaction: n(Fe) . n(O₂) = 4 : 3.

For 1.431moles of oxygen we need:

1.431 mol : n(Fe) = 3 : 4.

n(Fe) = 1.908 mol, there is no enough iron.

8. Answer is: there are 0.435 moles of C₆H₁₄.

N(C₆H₁₄) = 2.62·10²³; number of molecules.

n(C₆H₁₄) = N(C₆H₁₄) ÷ Na.

n(C₆H₁₄) = 2.62·10²³ ÷ 6.022·10²³ 1/mol.

n(C₆H₁₄) = 0.435 mol; amount of substance of C₆H₁₄.

Na - Avogadro constant or Avogadro number.

9. Answer is: 3.675 moles of carbon(II) oxide are required to completely react.

Balanced chemical reaction: Fe₂O₃(s) + 3CO(g) ⟶ 2Fe(s) + 3CO₂(g).

n(Fe₂O₃) = 1.225 mol; amount of substance.

From chemical reaction: n(Fe₂O₃) : n(CO) = 1 : 3.

1.225 mol : n(CO) = 1 : 3.

n(CO) = 3 · 1.225 mol.

n(CO) = 3.675 mol.

10. Answer is: there are 2.158 moles of barium atoms.

N(Ba) = 2.62·10²³; number of atoms of barium.

n(Ba) = N(Ba) ÷ Na.

n(Ba) = 1.3·10²⁴ ÷ 6.022·10²³ 1/mol.

n(Ba) = 2.158 mol; amount of substance of barium.

Na - Avogadro constant or Avogadro number.

11. Answer is: 6.26·10²³ carbon atoms are present.

n(C₂H₆O) = 0.52 mol; amount of substance.

N(C₂H₆O) = n(C₂H₆O) · Na.

N(C₂H₆O) = 0.52 mol · 6.022·10²³ 1/mol.

N(C₂H₆O) = 3.13·10²³.

In one molecule of C₂H₆O there are two atoms of carbon:

N(C) = N(C₂H₆O) · 2.

N(C) = 3.13·10²³ · 2.

N(C) = 6.26·10²³.

12. Answer is: the empirical formula is C₂H₄O.

If we use 100 grams of compound:

1) ω(C) = 51% ÷ 100% = 0.51.

m(C) = ω(C) · m(compound).

m(C) = 0.51 · 100 g.

m(C) = 51 g.

n(C) = m(C) ÷ M(C).

n(C) = 51 g ÷ 12 g/mol.

n(C) = 4.25 mol.

2) ω(H) = 9.3 % ÷ 100% = 0.093.

m(H) = 0.093 · 100 g.

m(H) = 9.3 g.

n(H) = 9.3 g ÷ 1 g/mol.

n(H) = 9.3 mol

3) ω(O) = 39.2 % ÷ 100%.

ω(O) = 0.392.

m(O) = 0.392 · 100 g.

m(O) = 39.2 g.

n(O) = 39.2 g ÷ 16 g/mol.

n(O) = 2.45 mol.

4) n(C) : n(H) : n(O) = 4.25 mol : 9.3 mol : 2.45 mol /2.45 mol.

n(C) : n(H) : n(O) = 1.73 : 3.795 : 1.