How many grams of lead(II) nitrate must be dissolved in 1.00 L of water to produce a solution that is 0.300 M in total dissolved
1 answer:
Answer:
A solution that is 0.100 M (i.e. 33,12 g of lead (II) nitrate in 1.00 L of water) yields the desired total dissolved ions concentration.
Explanation:
The molecular formula of lead (II) nitrate is Pb(NO and its molecular mass is 331,2 g/mol.
In disolution, the equilibrium will look like this:
Pb(NO ->
The equation above means that, one mol of lead (II) nitrate dissolved in 1L will yield one mol of Pb ions and 2 moles of NO3 ions, i.e. 3 moles total.
If we dissolve 0.100 moles of lead (II) nitrate in 1.00 L of water, the stoichiometry of the disolution states that in turn, it will yield 0.100 of Pb ions and 0.200 moles of NO3 ions, i.e. 0.300 M in total dissolved ions.
331,2 g/mol * 0.100 mol/L * 1 L = 33,12 grams of the compound are required.
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Answer:
1) Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.
2) Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.
Explanation:
1) Minimum mass of methane required to raise the temperature of water by 21.0°C.
Mass of water = m = 45.0 g
Specific heat capacity of water = c = 4.18 J/g°C
Change in temperature of water = ΔT = 21.0°C.
Heat required to raise the temperature of water by 21.0°C = Q
Q = 3,950.1 J = 3.9501 kJ
According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.
Then 3.950.1 kJ of heat will be given by:
Mass of 0.004923 moles of methane :
0.004923 mol × 16 g/mol=0.0788 g
Minimum mass of methane required to heat 45.0 g of water by 21.0°C is 0.0788 g.
2) Minimum mass of methane required to raise the temperature of water by 26.0°C.
Mass of water = m = 50.0 g
Specific heat capacity of water = c = 4.18 J/g°C
Change in temperature of water = ΔT = 26.0°C.
Heat required to raise the temperature of water by 21.0°C = Q
Q = 5,434 J= 5.434 kJ
According to reaction 1 mole of methane on combustion gives 802.3 kJ of heat.
Then 5.434 kJ of heat will be given by:
Mass of 0.006773 moles of methane :
0.006773 mol × 16 g/mol= 0.108 g
Minimum mass of methane required to heat 50.0 g of water by 26.0°C is 0.108 g.