Answer:
464.1 J absorbed.
Explanation:
Given data:
Specific heat of zinc = 0.39 J/g°C
Mass of zinc = 34 g
Temperature changes = 22°C to 57°C
Energy absorbed or released = ?
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 = 57°C - 22°C
ΔT = 35°C
Q = m.c. ΔT
Q = 34 g. 0.39 J/g°C. 35°C
Q = 464.1 J
The volume of a gas that is required yo react with 4.03 g mg at STP is 1856 ml
calculation/
- calculate the moles of Mg used
moles=mass/molar mass
moles of Mg is therefore=4.03 g/ 24.3 g/mol=0.1658 moles
- by use of mole ratio of Mg:O2 from the equation which is 2:1
the moles 02=0.1679 x1/20.0829 moles
- at STP 1 mole of a gas= 22.4 l
0.0895 moles=? L
- =0.0895 moles x22.4 l/ 1 mole=1.8570 L
into Ml = 1.8570 x1000=1856 ml approximately to 1860
1 kg = 1000g
2.43 kg *1000g/1kg = 2430 g
Answer:
6.72 g
Explanation:
Given data:
Mass of calcium chlorate = 13.8 g
Mass of oxygen produced = ?
Solution:
Chemical equation:
Ca(ClO₃)₂ → CaCl₂ + 3O₂
Number of moles of calcium chlorate:
Number of moles = mass / molar mass
Number of moles = 13.8 g/ 206.98 g/mol
Number of moles = 0.07 mol
Now we will compare the moles of oxygen and calcium chlorate.
Ca(ClO₃)₂ : O₂
1 : 3
0.07 : 3×0.07=0.21 mol
Mass of oxygen:
Mass = number of moles × molar mass
Mass = 0.21 mol × 32 g/mol
Mass = 6.72 g