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 balanced chemical reaction would be :
S + O2 = SO2
2SO2 + O2 = 2SO3
We are given the amount of oxygen to be used in the reaction. This will be the starting value in the calculations. We do as follows:
2.56 L O2 ( 1 mol / 22.4 L) ( 1 mol SO2 / 1 mol O2 ) ( 2 mol SO3 / 2 mol SO2 ) 22.4 L / 1 mol = 2.56 mol SO3
The concentration of the solution is 4.25 M
Explanation
molarity=moles/volume in liters
moles = mass/molar mass
molar mass of HF = 19 + 1 = 20 g/mol
moles is therefore = 17.0 g/ 20 g/mol = 0.85 moles
volume in liters = 2 x10^2ml/1000 = 0.2 liters
therefore molarity = 0.85/0.2 = 4.25 M
On the bottom
1. sharing electrons
2. the middle two dots between the F:F
