The percentage of water of crystallization in blue vitriol is 36.07%.
M(H₂O) = 2Ar(H) + Ar(O) x g/mol
M(H₂O) = 2 + 16 x g/mol
M(H₂O) = 18 g/mol; molar mass of water
M(CuSO₄·5H₂O) = Ar(Cu) + Ar(S) + 4Ar(O) + 5Mr(H₂O) x g/mol
M(CuSO₄·5H₂O) = 63.5 + 32 + 64 + 90 x g/mol
M(CuSO₄·5H₂O) = 249.5 g/mol; molar mass of copper sulphate pentahydrate
The percentage of water: 5M(H₂O) / M(CuSO₄·5H₂O) x 100%
The percentage of water: 90 g/mol / 249.5 g/mol x 100%
The percentage of water = 36.07%
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Answer:
I can't draw but you could draw 2 electrons in the first orbit and 3 electrons in the second orbit.
Explanation:
Answer:
c = 0.898 J/g.°C
Explanation:
1) Given data:
Mass of water = 23.0 g
Initial temperature = 25.4°C
Final temperature = 42.8° C
Heat absorbed = ?
Solution:
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
Specific heat capacity of water is 4.18 J/g°C
ΔT = 42.8°C - 25.4°C
ΔT = 17.4°C
Q = 23.0 g × × 4.18 J/g°C × 17.4°C
Q = 1672.84 j
2) Given data:
Mass of metal = 120.7 g
Initial temperature = 90.5°C
Final temperature = 25.7 ° C
Heat released = 7020 J
Specific heat capacity of metal = ?
Solution:
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 = 25.7°C - 90.5°C
ΔT = -64.8°C
7020 J = 120.7 g × c × -64.8°C
7020 J = -7821.36 g.°C × c
c = 7020 J / -7821.36 g.°C
c = 0.898 J/g.°C
Negative sign shows heat is released.
Answer:
See Explanation
Explanation:
Hence the mass defect is;
[235.04393 + 1.00867] - [ 136.92532 + 96.91095 + 2(1.00867)]
= 236.0526 - 235.85361
= 0.19899 amu
Since 1 amu = 1.66 * 10^-27 Kg
0.19899 amu = 0.19899 * 1.66 * 10^-27 = 3.3 * 10^-28 Kg
Binding energy = Δmc^2
Binding energy = 3.3 * 10^-28 Kg * (3 * 10^8)^2 = 2.97 * 10^-11 J
ii) 
Hence the mass defect is;
[10.01294 + 1.00867] - [7.01600 + 4.00260]
= 11.02161 - 11.0186
= 0.00301 amu
Since 1 amu = 1.66 * 10^-27 Kg
0.00301 amu = 0.00301 * 1.66 * 10^-27 = 4.997 * 10^-30 Kg
Binding energy = Δmc^2
Binding energy = 4.997 * 10^-30 Kg * (3 * 10^8)^2 = 4.5 * 10^-13 J
