Answer:
Final temperature is 34.2 °C
Explanation:
Given data:
mass of metal = 125 g
temperature of metal = 93.2 °C
mass of water v= 100 g
temperature of water = 18.3 °C
specific heat of meta is = 0.900 j/g. °C
specific heat of water is = 4.186 j/g. °C
final temperature of water and metal = ?
Solution:
Q = m . c . ΔT
ΔT = T2-T1
now we will put the values in equation
Q1 = m . c . ΔT
Q1 = 125 g. 0.900 j/g. °C .93.2°C - T2
Q1 = 112.5 (93.2°C - T2)
Q1 =10,485 - 112.5T2
Q2 = m . c . ΔT
Q2 = 100 . 4.186. (T2- 18.3)
Q2 = 418.6 . (T2- 18.3)
Q2 = 418.6T2 - 7660.38
10,485 - 112.5T2 = 418.6T2 - 7660.38
10,485 + 7660.38 = 418.6T2+ 112.5T2
18145.38 = 531.1 T2
T2 = 18145.38/531.1
T2 = 34.2 °C
The Lewis dot diagram structure for hydrogen cyanide would be the following:
H - C --- N
--- = triple bond.
