½H2(g) + ½I2(g) → HI(g) ΔH = +6.2 kcal/mol
or...
½H2(g) + ½I2(g) + 6,2kcal/mole → HI(g)
________
21.0 kcal/mole + C(s) + 2S(s) → CS2(l)
or...
C(s) + 2S(s) → CS2(l) ΔH = +2,1 kcal/mole
_________
ΔH > 0 ----------->>> ENDOTHERMIC REACTIONS
Answer:
I think A it looks to be the answer
Explanation:
sorry if wrong
Answer: The concentration of the OH-, CB = 0.473 M.
Explanation:
The balanced equation of reaction is:
2HCl + Ca(OH)2 ===> CaCl2 + 2H2O
Using titration equation of formula
CAVA/CBVB = NA/NB
Where NA is the number of mole of acid = 2 (from the balanced equation of reaction)
NB is the number of mole of base = 1 (from the balanced equation of reaction)
CA is the concentration of acid = 1M
CB is the concentration of base = to be calculated
VA is the volume of acid = 23.65 ml
VB is the volume of base = 25mL
Substituting
1×23.65/CB×25 = 2/1
Therefore CB =1×23.65×1/25×2
CB = 0.473 M.
