The student needs to dilute the stock to get the proper concentration. And during the dilution, the molar number of solute will not change. So the volume of stock needed is 2*0.1/1.75=0.114 L. Then dilute to 2.00 L.
A. MnS(s) + 2H + (aq) > Mn2 + (aq) + H2S(g)
0,35 kmol/m³ = 0,35 mol/dm³ = 0,35 mol/L
175 mL = 0,175 L
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C = n/V
n = 0,35×0,175
n = 0,06125 mol
mCa(NO₃)₂: 40+(14×2)+(16×6) = 164 g/mol
1 mol --------- 164g
0,06125 ---- X
X = 10,045g
To prepare 175 mL of 0,35M solution, add 10,045g of calcium nitrate and add water to a volume of 175ml.
Explanation is in a file
bit.ly/3a8Nt8n