1:2:6? Ba(NO3)2 +Na2SO4 -----> BaSO4 + 2NaNO3
Ba(NO3)2
BaSO4
NaNO 3
Na2SO4
Answer:
Specific heat of alloy = 0.2 j/ g.°C
Explanation:
Specific heat capacity:
It is the amount of heat required to raise the temperature of one gram of substance by one degree.
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
Given data:
Mass of bold = 25 g
Heat absorbed = 250 J
Initial Temperature = 25°C
Final temperature = 78°C
Specific heat of alloy = ?
Solution:
Change in temperature:
ΔT = 78°C - 25°C
ΔT = 53°C
Now we will put the values in formula.
Q = m.c. ΔT
250 j = 25 g × c ×53°C
250 j = 1325 g.°C × c
250 j / 1325 g.°C = c
c = 0.2 j/ g.°C
The formula unit for Li and O is —————-> Li2O
It is greater than the total mass
