Question Library

2H2 + O2 → 2H2O 2V3 1V3 2V 50cm 100 — 50cm 25cm3 50cm3 Volume of gas left = (100 - 25)cm3 of O2 = 75 cm3 of O2

H2 = 20cm O2 = 35cm N2 = 10cm CO = 15cm Total vol = 80cm Mole fraction of H2 = (20)/(80) = 0.25

Hydrocarbon means the compound comprises Hydrogen and Carbon atoms. % of H\(_2\) = 14.29 % of C = 100 - 14.29 = 85.71 Solving the empirical formula for the Hydrocarbon, Atom of element C H % by mass 85.71 14.29 Divide by their Atomic masses \(\frac{85.71}{12}\) \(\frac{14.29}{1}\) Divide through by the smaller number \(\frac{7.1425}{7.1425}\) \(\frac{14.29}{7.1425}\) Ratio 1 : 2 Empirical formula: CH\(_2\) Recall that (Empirical formula)x n = Molecular mass (CH\(_2\))n = Molecular mass We need to get the molecular mass.To do that, we were given the: mass, m = 0.07g and V = 56 cm\(^3\) But Number of moles = \(\frac{volume}{Molar Volume}\) = \(\frac{56}{22,400}\) = 0.0025 mole N.B : At stp, Molar volume of all gases = 22.4dm\(^3\) or 22,400cm\(^3\) Thus, n = \(\frac{m}{M}\) M = \(\frac{m}{n}\) = \(\frac{0.07}{0.0025}\) = 28. (CH\(_2\))n = Molecular mass (12 + (1 x 2))n = 28 14n = 28 n = \(\frac{28}{14}\) = 2 The formula for the hydrocarbon = (CH\(_2\))n = (CH\(_2\))\(_2\) = C\(_2\)H\(_4\)

P1 = 750MHg V1 = 100cm3 l2 = 308k P2 = 100 mmHg V2 = ? But (P1V1)/(T1) = ((P2V2)/(T2) i.e (750 x 100)/(308) = (1000 x )/(V2)/(308) V2 = (750 x 1000 x 308)/(308 x 1000) = 75 cm3

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option E.

No expanded working has been attached for this question yet. The verified answer is Option D.

Cu\(^{2+}\) + 2e- → Cu 1mole of electron = 1F = Number of moles of Cu ions 2moles = 2F = 1 mole of Cu 2 x 96500C = I mole of Cu 10 x 3600C = x moles X = \(\frac{36000}{193000}\) X = 0.186 mole of Cu dissolved from the anode. Also, during electrolysis, copper ions are reduced to copper metals at the cathode(negative electrode). This, however, decreases the concentration of Cu ions present in the solution due to the reduction reaction at the cathode. These make option C the correct answer.

No expanded working has been attached for this question yet. The verified answer is Option C.

No expanded working has been attached for this question yet. The verified answer is Option B.

No expanded working has been attached for this question yet. The verified answer is Option B.

CH\(_3\) - C≡CH + 2Br\(_2\) → CH\(_3\)-CBr\(_2\) -CH(Br\(_2\)) 1 mole : 2 moles Recall that n = \(\frac{m}{M}\) Molar mass of Br\(_2\) = 80 x 2 = 160 g/mol n = \(\frac{8}{160}\) n = 0.05 mole If 1 mole of CH\(_3\) - C≡CH reacts with 2 moles of Br\(_2\) then X mole of CH\(_3\) - C≡CH will react with 0.05 mole of Br\(_2\) X = \(\frac{0.05}{2}\) X = 0.025 mole of CH\(_3\) - C≡CH Also, recall that n = \(\frac{m}{M}\) Molar mass of CH\(_3\) - C≡CH = C3H4 = (12 X 3) + (1 X 4) = 40 g/mol n = \(\frac{m}{M}\) n = 0.025 x 40 = 1.0 g

Given: Volume, V = 500cm\(^3\) Concentration , C = 0.2 M n = \(\frac{CV}{1000}\) n = \(\frac{{0.2}\times{500}}{1000}\) n = 0.1 mole Also, recall that n = \(\frac{mass}{Molar mass}\) = \(\frac{m}{M}\) m = n X M Molar mass of NaOH = 23 + 16 + 1 = 40g/mol m = 0.1 x 40 = 4g

Adding a catalyst increases the rate of a chemical reaction. A catalyst is a substance that speeds up a reaction without being used up in the process. Also, increasing the surface area of a reactant increases the rate of a chemical reaction. This is because it increases the number of collisions between the reactants. Similarly, increasing the temperature also affects the rate of chemical reaction. Therefore, Increase in temperature, increase in concentration, adding a catalyst and increase in surface area affect the rate of chemical reaction.

Ca(HCO2) → CaCO3 + H2O + CO2 is not a redox reaction