Answer:
Explanation:
250 cm^3 of 0.2 moldm-3 H2SO4 can be prepared from 150cm^3 of 1.0 moldm^-3 by dilution.
150cm^3 of the 1.0 moldm^-3 stock solution is measured out using a measuring cylinder and transferred into a 250 cm^3 standard volumetric flask and made up to mark. The resulting solution is now 250cm^3 of 0.2 moldm-3 H2SO4.
Answer:
(a) 13.7 g.
(b) 28.91 g.
Explanation:
- molality (m) is the no. of moles of solute dissolved in 1.0 kg of solvent.
∴ m = (no. of moles of solute)/(mass of water (kg))
<em>∴ m = (mass/molar mass of solute)/(mass of water (kg)).</em>
<em />
<u><em>(a) Calculate the mass of CaCl₂·6H₂O needed to prepare 0.125 m CaCl₂(aq) by using 500. g of water.</em></u>
∵ m = (mass/molar mass of CaCl₂·6H₂O)/(mass of water (kg)).
m = 0.125 m, molar mass of CaCl₂·6H₂O = 219.0757 g/mol, mass of water = 500.0 g = 0.5 kg.
∴ 0.125 m = (mass of CaCl₂·6H₂O / 219.0757 g/mol)/(0.5 kg).
∴ mass of CaCl₂·6H₂O = (0.125 m)(219.0757 g/mol)(0.5 kg) = 13.7 g.
<u><em>(b) What mass of NiSO₄·6H₂O must be dissolved in 500. g of water to produce 0.22 m NiSO₄(aq)?</em></u>
∵ m = (mass/molar mass of NiSO₄·6H₂O)/(mass of water (kg)).
m = 0.22 m, molar mass of NiSO₄·6H₂O = 262.84 g/mol, mass of water = 500.0 g = 0.5 kg.
∴ 0.125 m = (mass of NiSO₄·6H₂O / 262.84 g/mol)/(0.5 kg).
∴ mass of NiSO₄·6H₂O = (0.22 m)(262.84 g/mol)(0.5 kg) = 28.91 g.
I’m guessing it would be A because the river discharge tends to gather near coastlines where the river ends
10 atoms in total
4 atoms of H
4 atoms of O
2 atoms of Zn
Hope this helps :)
