Answer:
The weight of 3.45 moles of carbon dioxide has been 151.8 grams.
Moles can be calculated as the mass of solute present with respect to the molecular mass of the solute.
Moles can be expressed as:
Moles = \rm \dfrac{weight}{molecular\;weight}
molecularweight
weight
The molecular weight of carbon dioxide has been 44 grams/mol.
The given moles of carbon dioxide = 3.45 moles.
3.45 mol = \rm \dfrac{weight}{44\;g/mol}
44g/mol
weight
Weight of carbon dioxide = 3.45 \times× 44 grams
Weight of carbon dioxide = 151.8 grams.
The weight of 3.45 moles of carbon dioxide has been 151.8 grams.
The elements are oxygen and carbon
Answer:
-767,2kJ
Explanation:
It is possible to sum enthalpies of half-reactions to obtain the enthalpy of a global reaction using Hess's law. For the reactions:
1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁= −241.8 kJ
2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂= +361.7 kJ
3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl (g) ΔH₃= −92.3 kJ
4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄= − 607.9 kJ
5) H₂O(g) ⟶ H₂O(l) ΔH₅= − 44.0 kJ
The sum of (4) - (2) produce:
6) XCl₄(s) + O₂(g) ⟶ XO₂(s) + 2Cl₂(g) ΔH₆ = ΔH₄ - ΔH₂ = -969,6 kJ
(6) + 4×(3):
7) XCl₄(s) + 2H₂(g) + O₂(g) ⟶ XO₂(s) + 4HCl(g) ΔH₇ = ΔH₆ + 4ΔH₃= -1338,8 kJ
(7) - 2×(1):
8) XCl₄(s) + 2H₂O(g) ⟶ XO₂(s) + 4HCl(g) ΔH₈ = ΔH₇ - 2ΔH₁= -855,2kJ
(8) - 2×(5):
9) XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g) ΔH₉ = ΔH₈ - 2ΔH₅= <em>-767,2kJ</em>
I hope it helps!
Answer:
Number of moles = 3.82 mol
Explanation:
Given data:
Number of moles of CO₂ = ?
Mass of CO₂ = 168.2 g
Solution:
Number of moles = mass/molar mass
Molar mass of CO₂ = 44 g/mol
By putting values,
Number of moles = 168.2 g/ 44 g/mol
Number of moles = 3.82 mol
1.7960L
Explanation:
the mass of the gas is constant in both instances
pv/T=constant(according to pv=nRT)
745mmHg*2L/298K=760mmHg*v/273K
v=1.7960L