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A sample of calamine ore (ZnCO3) contains clay as an impurity which is 37.5%. Clay impurity loses some amount of its weight as moisture on prolonged heating. If the total loss in weight of ore on heating (due to CO2 and H2O) is 28%, then find the percentage loss in the weight of the clay. (Given atomic masses: Zn=65,C=12,O=16).
A sample of calamine ore (ZnCO3) contains clay as impurity which is 37.5%. Let the total mass of the sample be 100g ∴mZnCO3=62.5g and mclay=37.5g ZnCO3→ZnO+CO2. 125g of ZnCO3 corresponds to 44g of CO2. ∴62.5g corresponds to 22g of CO2. mH2O+mCO2=28g ∴mH2O=28−22=6g ∴% loss in the weight of clay =637.5×100=16%
2.76 g of silver carbonate on being strongly heated yields a residue weighing:
When silver carbonate is strongly heated, a residue of silver is obtained. The reaction is as follows: Ag2CO3(s)⟶Heat2Ag(s)+CO2(g)+12O2(g) The molecular weight of silver carbonate and silver is 276g/mol and 108g/mol respectively. Thus, when 276g (or 1mole ) of silver carbonate is heated, 2×108=216g(2moles) of silver is obtained. When 2.76g of silver carbonate is heated, the amount of residue (silver) obtained is 216276×2.76=2.16g.
We have two solutions, one is 1 L of 0.1 M NaCl and the other is 2 L of 0.2 M CaCl2. Using only these two solutions, what maximum volume of a solution can be prepared, which has [Cl−]=0.34M?
Given: Both electrolytes are strong.
o.i MNaCl NaCl→Na++Cl− I mol of NaCl=1 mol of Cl In 1 ltr solution, o.t mol of NaCl =0.1mol of Cl− 0.2MCaCl2−2ltr CaCl2→Ca++2Cl− I mol of CaCl2=2 mol of Cl− In 1 ler solution, 0.2 mol of CaCl2=0.2×2mol of Cl− In 2 ltr solution =0.8 mol of Cl− Total moles of Cl− =00.8+0.1=0.9mol Desired Molarity of Cl−=0.34M So, volume of solution = 0.9mol0.34M=2.65 litr
There are two oxides of sulphur. They contain 50 % and 60% of oxygen, respectively, by weights. The ratio of weights of sulphur, which combines with 1 g of oxygen, is:
In the first oxide, the percentages of O and S are 50% and 50% respectively. Thus, 1g of oxygen will combine with 1g of sulphur. In the second oxide, the percentages of O and S are 60% and 40%, respectively. Thus, 1g of oxygen will combine with 23g of sulphur. The ratio of weights of sulphur which combine with 1 g of oxygen =1:23=3:2
Na2S2O3→Na2SO4 The total change in oxidation number =4×2=8 ∴ENa2S2O3=mol.wtV.f=M8
The balanced reaction is CaCO3(s)+2HCl(aq)→CaCl2(aq)+CO2(g)+H2O(l)10g calcium carbonate corresponds to 10100=0.1mole (Molecular weight of CaCO3=100g/mol). It will react with 0.2molHCl. However, 0.3molHCl is present. Hence, calcium carbonate is the limiting reagent. 0.1mol of calcium carbonate will form 0.1mol of calcium chloride.
Find the mass of pure hydrogen chloride gas dissolved per litre of its solution if the normality of the solution is 0.15N. [At. Wt. : H=1,Cl=35.5]
The solution is 0.15N. Hydrogen chloride (HCl) contains one replaceable H atom. Molar mass of HCl=(1+35.5)=36.5u The expression for the equivalent weight of an acid is as given below. Equivalent weight=Molecular mass/Number of replaceable H HCl contains one replacable hydrogen. Substitute values in the above expression. =36.51=36.5u The expression for the normality of solution is as given below. =Solute in grams per litre of solution/Equivalent weight in grams Solute in grams per litre of solution = Normality Equivalent weight in grams. The mass of pure hydrogen chloride gas dissolved per litre of its solution =(0.15×36.5)g=5.475g/l
45.4 L of dinitrogen reacted with 22.7L of dioxygen and 45.4 L of nitrous oxide was formed. The reaction is given below:
2N2(g)+O2(g)→2N2O(g)
Which law is being obeyed in this experiment?
The ratio of the volumes of dinitrogen, oxygen and nitrous oxide is 45.4:22.7:45.4. This is in the simple whole number ratio 2 : 1 : 2 This satisfies the Gay Lussac's law of combining volume of gases.
The balanced reaction is given below: 2SO2+O2+2H2O→2H2SO4 5.6 4.8 (No. of moles given) 5.624.81 (Ratio of moles given and stoichiometry coefficient) =2.8=4.8 As 2.8 is lower than 4.8,SO2 is the limiting reagent and product will form according to number of moles of SO2 (given). So, moles produced of H2SO4=22×5.6=5.6mol
The mass percent of calcium, phosphorus and oxygen in calcium phosphate, Ca3(PO4)2 is :
Molecular formula of calcium phosphate is Ca3(PO4)2 Its molar mass is 3(40)+2(31)+8(16)=310g/mol Percent of calcium =120310×100=38.71% Percent of phosphorus =62310×100=20.0% Percent of oxygen =128310×100=41.29%
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