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 is: ph value is 3.56.
Chemical reaction 1: H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq); Ka₁ = 4,3·10⁻⁷.
Chemical reaction 2: HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5,6·10⁻¹¹.
c(H₂CO₃) = 0,18 M.
[HCO₃⁻] = [H⁺<span>] = x.
</span>[H₂CO₃] = 0,18 M - x.
Ka₁ = [HCO₃⁻] · [H⁺] / [H₂CO₃].
4,3·10⁻⁷ = x² / (0,18 M -x).
Solve quadratic equation: x = [H⁺] =0,000293 M.
pH = -log[H⁺] = -log(0,000293 M).
pH = 3,56; second Ka do not contributes pH value a lot.
Explanation:
no of moles = no of atoms ÷ avogadro's number
= (9.8×10^24) ÷ (6.02×10^23)
<h2>
Answer:</h2><h2>
The percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months = 19.4 %</h2>
Explanation:
Average electricity consumed per month = 900 kWh
The family cools their house for three months during the summer with two window-unit air conditioners
The power consumed by one window-unit air conditioners = 350 kWh
The power consumed by two window-unit air conditioners = 350(2) = 700 kWh
Power consumed for two air conditioners for the three summer months = 700 (3) = 2100 kWh
Total power consumed for 1 year = 900 (12) = 10800kWh
The percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months = 
= 19.4 %
Answer:
Q = 2640.96 J
Explanation:
Given data:
Mass of He gas = 10.7 g
Initial temperature = 22.1°C
Final temperature = 39.4°C
Heat absorbed = ?
Solution:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree. Specific heat capacity of He is 14.267 J/g.°C
Formula:
Q = m.c. ΔT
Q = amount of heat absorbed or released
m = mass of given substance
c = specific heat capacity of substance
ΔT = change in temperature
ΔT = 39.4°C - 22.1°C
ΔT = 17.3°C
Q = 10.7 g× 14.267 J/g.°C × 17.3°C
Q = 2640.96 J
