Answer:
48.37514 kj
Explanation:
Given data:
Mass of water = 163 g
Initial temperature = 29°C
Final temperature = 100°C
Heat added = ?
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 water is 4.18 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 = Final temperature - initial temperature
ΔT = 100°C - 29°C
ΔT = 71°C
Q = 163 g × 4.18 j/g.°C × 71°C
Q = 48375.14 j
Joule to Kj conversion:
48375.14 /1000 = 48.37514 kj
The answer is c because ice is cold
Answer is: <span>the percent ionization is 0,19%.
</span>Chemical reaction: HA(aq) ⇄ H⁺(aq) + A⁻(aq).
Ka(HA) = 3,6·10⁻⁷.
c(HA) = 0,1 M.
[H⁺] = [A⁻] = x; equilibrium concentration.
[HA] = 0,1 M - x.
Ka = [H⁺] · [A⁻] / [HA].
0,00000036 = x² / 0,1 M - x.
Solve quadratic equation: x = 0,00019 M.
α = 0,00019 M ÷ 0,1 M · 100% = 0,19%.
The answer would for sure be 178 because acid & base make that amount so yea!! I’m talllyyy right
