No sorry because i'm only 12 years old so no
I got B. Because the answer is K2SO4, I got it right on my test so I know its right
It would be NaOH + HCl → <span>NaCl + H2O
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NaOH is sodium hydroxide, which is a strong base. HCl is hydrochloric acid, which is a strong acid.
You have a strong base and a strong acid on the left side, however, at the result side, you end up with NaCl + H2O. Sodium chloride is simply table salt and H2O is just water, thus it has been neutralized.
The precipitate that is most likely formed from a solution containing Ba+2, Li+, OH-1, and CO3^-2 is BaCO3.
This is because carbonates of all metals except sodium, Lithium potassium (group 1) and ammonium are insoluble in water. Hydroxides of sodium, Lithium, potassium and ammonium are very soluble in water, calcium and barium are moderately soluble. Ba(CO3) is insoluble in water and therefore forms a precipitate.
Answer:
the heat of formation of isopropyl alcohol is -317.82 kJ/mol
Explanation:
The heat of combustion of isopropyl alcohol is given as follows;
C₃H₇OH (l) +(9/2)O₂ → 3CO₂(g) + 4H₂O (g)
The heat of combustion of CO₂ and H₂O are given as follows
C (s) + O₂ (g) → CO₂(g) = −393.50 kJ
H₂ (g) + 1/2·O₂(g) → H₂O (l) = −285.83 kJ
Therefore we have
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ which we can write as
3C (s) + 3O₂ (g) → 3CO₂(g) = −393.50 kJ × 3 =
4H₂ (g) + 2·O₂(g) → 4H₂O (l) = −285.83 kJ × 4
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ = +2006 kJ/mol
-1180.5 - 1143.32 +2006 = -317.82 kJ/mol
Therefore, the heat of formation of isopropyl alcohol = -317.82 kJ/mol.
