Laws of Mass Conservation
Trending Questions
What is the equivalent weight of Calcium carbonate ?
- 11.65 g
- 23.3 g
- 25.5 g
- 30.6 g
P4+O2⟶P4O10
P4O10+H2O⟶H3PO4
(Molar mass of P = 31 g/mol)
- 150 g
- 288 g
- 560 g
- 392 g
- Conservation of mass
- Constant proportions
- Reciprocal proportions
- Multiple proportions
of Calcium carbonate on heating gave of Calcium oxide and of Carbon dioxide. Show that these observations are in agreement with law of conservation of mass.
Rate of reaction is proportional to product of molar concentration of reactants with each concentration term raised to power its stiochiometric coefficient. This is the law of
Mass action
Equilibrium
Constant proportion
Reciprocal proportions
According to the law of conservation of mass, the mass of the products is equal to the mass of the reactants.
- True
- False
- 12 g of carbon combines with 32 g of oxygen to form 44 g of carbon dioxide.
- When 12 g of carbon is heated in a vacuum, there is no change
in mass. - A sample of air increases in volume when heated at constant pressure but its mass remains unaltered.
- The weight of a piece of platinum is the same before and after heating in air.
(Molar mass of Ag = 108 g/mol, Cl = 35.5 g/mol, Br = 80 g/mol)
- 4.85%
- 9.64%
- 8.45%
- 7.23%
- 11 g
- 12 g
- 13 g
- 10 g
- 12 g of carbon combines with 32 g of oxygen to form 44 g of CO2
- When 12 g of carbon is heated in a vacuum there is no change in mass.
- The weight of a piece of platinum is the same before and after heating in air.
- A sample of air increases in volume when heated at constant pressure but its mass remains unaltered
- 6.5 g
- 5.6 g
- 4.2 g
- 5.3 g
- 0.0651 g
- 0.0782 g
- 0.0440 g
- 0.0200 g
SO2(g)+Cl2(g)+2H2O→H2SO4(l)+2HCl(l)
H2SO4(l)+BaCl2→BaSO4↓+2HCl
112 g of sulphur is burned initially to form SO2. Find the amount of BaSO4 precipitate obtained in the series of reactions.
(Molar mass of Ba = 137 g/mol)
- 815.5 g
- 456.3 g
- 635.7 g
- 988.1 g
CuSO4.5H2O→CuSO4+5H2O
CuSO4+2KCN→Cu(CN)2+K2SO4
2Cu(CN)2→2CuCN+(CN)2
2CuCN+6KCN→2K3[Cu(CN)4]
(Molar mass of K=39 g/mol, Cu=63.5 g/mol)
- 122.4 g
- 19.96 g
- 12.24 g
- 49.9 g
- Conservation of mass
- Constant proportions
- Reciprocal proportions
- Multiple proportions
(Molar mass of K = 39 g/mol, Fe = 56 g/mol, Zn = 65 g/mol)
- 15.85 g
- 10.67 g
- 6.35 g
- 20.74 g
SO2(g)+Cl2(g)+2H2O→H2SO4(l)+2HCl(l)
H2SO4(l)+BaCl2→BaSO4↓+2HCl
112 g of sulphur is burned initially to form SO2. Find the amount of BaSO4 precipitate obtained in the series of reactions.
(Molar mass of Ba = 137 g/mol)
- 815.5 g
- 456.3 g
- 635.7 g
- 988.1 g
Five grams of KClO3 yield 3.041 g of KCl and 1.36 L of oxygen at standard temperature and pressure. Do these figures support the law of conservation of mass within limits of ±0.4% error?
True
False