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:
pH = 2.66
Explanation:
- Acetic Acid + NaOH → Sodium Acetate + H₂O
First we <u>calculate the number of moles of each reactant</u>, using the <em>given volumes and concentrations</em>:
- 0.75 M Acetic acid * 50.0 mL = 37.5 mmol acetic acid
- 1.0 M NaOH * 10.0 mL = 10 mmol NaOH
We<u> calculate how many acetic acid moles remain after the reaction</u>:
- 37.5 mmol - 10 mmol = 27.5 mmol acetic acid
We now <u>calculate the molar concentration of acetic acid after the reaction</u>:
27.5 mmol / (50.0 mL + 10.0 mL) = 0.458 M
Then we <u>calculate [H⁺]</u>, using the<em> following formula for weak acid solutions</em>:
- [H⁺] =

Finally we <u>calculate the pH</u>:
Answer:
(B) II, IV.
hope this answer is helpful for u.
<span>C. Carbon. H. Hydrogen. N. Nitrogen. O. Oxygen. P. Phosphorus. <span>S. Sulfur.</span></span>