You can pick it up and move it
Problem One
You will use both m * c * deltaT and H = m * heat of fusion.
Givens
m = 12.4 grams
c = 0.1291
t1 = 26oC
t2 = 1204
heat of fusion (H_f) = 63.5 J/grams.
Equation
H = m * c * deltaT + m * H_f
Solution
H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5
H = 1660.1 + 787.4
H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.
Problem Two
Formula and Givens
t1 = 14.5
t2 = 50.0
E = 5680
c = 4.186
m = ??
E = m c * deltaT
Solution
5680 = m * 4.186 * (50 - 14.5)
5680 = m * 4.186 * (35.5)
5680 = m * 148.603 * m
m = 5680 / 148.603
m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.
m = 38.2 to 3 sig digs.
Answer:
HOAc is stronger acid than HClO
ClO⁻ is stronger conjugate base than OAc⁻
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
Explanation:
Assume 0.10M HOAc => H⁺ + OAc⁻ with Ka = 1.8 x 10⁻⁵
=> [H⁺] = √Ka·[Acid] =√(1.8 x 10⁻⁵)(0.10) M = 1.3 x 10⁻³M H⁺
Assume 0.10M HClO => H⁺ + ClO⁻ with Ka = 3 x 10⁻⁸
=> [H⁺] = √(3 x 10⁻⁸)(0.10)M = 5.47 x 10⁻⁵M H⁺
HOAc delivers more H⁺ than HClO and is more acidic.
Kb = Kw/Ka, Kw = 1 x 10⁻¹⁴
Kb(OAc⁻) = 5.5 x 10⁻¹⁰
Kb(ClO⁻) = 3.3 x 10⁻⁷
Answer:
Mass = 15.20 g of KCl
Explanation:
The balance chemical equation for the decomposition of KClO₃ is as follow;
2 KClO₃ = 2 KCl + 3 O₂
Step 1: Calculate moles of KClO₃ as;
Moles = Mass / M/Mass
Moles = 25.0 g / 122.55 g/mol
Moles = 0.204 moles
Step 2: Find moles of KCl as;
According to equation,
2 moles of KClO₃ produces = 2 moles of KCl
So,
0.204 moles of KClO₃ will produce = X moles of KCl
Solving for X,
X = 2 mol × 0.204 mol / 2 mol
X = 0.204 mol of KCl
Step 3: Calculate mass of KCl as,
Mass = Moles × M.Mass
Mass = 0.204 mol × 74.55 g/mol
Mass = 15.20 g of KCl
