D. Glucose it’s is stored in the form of glucose inside of chloroplasts
1 mole ----------- 22.4 L ( at STP )
3.75 moles ----- ?
3.75 x 22.4 / 1 => 84.0 L
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!
0,15 moles of NaOH-------in------------1000ml
x moles of NaOH------------in--------100ml
x = 0,015 moles of NaOH
final volume = 150ml
0,015 moles of NaOH---in-------150ml
x moles of NaOH--------------in-----1000ml
x = 0,1 moles of NaOH
answer: 0,1mol/dm³ (molarity)
