Answer:
Q = 30355.2 J
Explanation:
Given data:
Mass of ice = 120 g
Initial temperature = -5°C
Final temperature = 115°C
Energy required = ?
Solution:
Specific heat capacity of ice is = 2.108 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
Q = m.c. ΔT
ΔT = T2 -T1
ΔT = 115 - (-5°C)
ΔT = 120 °C
Q = 120 g × 2.108 j/g.°C × 120 °C
Q = 30355.2 J
207.217 amu
Work:
203.973 amu *(0.014) = 2.855 amu
205.974 amu *(0.241) = 49.639 amu
206.976 amu *(0.221) = 45.741 amu
207.977 amu *(0.524) = 108.979 amu
2.855 + 49.639 + 45.741 + 108.979 = 207.217 amu
When the atomic number increases The atomic size will be larger.
Answer:
7.28
Explanation:
-log(5.2x10^-8) = 7.28
- Hope that helped! Please let me know if you need further explanation.
