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.
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.
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.