A.) <span>Each molecule of hydrochloric acid, HCl, contains one atom of hydrogen and "One atom of Chlorine"
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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!
1) in periodic acid (HIO₄), iodine has oxidation number +7, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 4 · (-2) = 0.
x = +7.
2) in molecule of iodine (I₂), iodine has oxidation number 0, because iodine is nonpolar molecule.
3) in sodium iodide (NaI), iodine has oxidation number -1, sodium has oxidation number +1:
+1 + x = 0.
x = -1.
4) in iodic acid (HIO₃), iodine has oxidation number +5, hydrogen has oxidation number +1, oxygen has -2, compound has neutral charge:
+1 + x + 3 · (-2) = 0.
x = +5.
Answer:
38.541 × 10¹⁹ formula units
Explanation:
Given data:
Mass of chromium sulfate = 0.25 g
Formula units in 0.25 g = ?
Solution:
The given problem will solve by using Avogadro number.
It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.
The number 6.022 × 10²³ is called Avogadro number.
For example,
18 g of water = 1 mole = 6.022 × 10²³ formula units of water
Number of moles of chromium sulfate = Mass / molar mass
Number of moles of chromium sulfate = 0.25 g/ 392.16 g/ mol
Number of moles of chromium sulfate = 6.4 × 10⁻⁴ moles
Number of formula units:
1 mole = 6.022 × 10²³ formula units
6.4 × 10⁻⁴ moles × 6.022 × 10²³ formula units / 1 mol
38.541 × 10¹⁹ formula units