NSMQ Past Questions; Chemistry (02 and 03) pdf.
NSMQ Past Questions; Chemistry (02 and 03) pdf.
Preamble to all schools:
Consider the following elements, listed not in any particular order.
Beryllium, Argon, Silicon, and Sodium.
- Which of these elements has the lowest first ionization energy?
Ans: Sodium
- Which of these elements has the biggest atomic radius?
Ans: Argon
- Which of these elements is most electronegative?
Ans: Silicon
-
Why was VALCO, the first aluminum smelter company in Ghana, sited at Tema in the Greater Accra Region.
Ans: Sited at Tema so as to receive imported alumina through the Tema harbour and also having electricity supply from Akosombo.
- Why is the electrolysis of molten alumina preferred to electrolysis of an aqueous solution of Al3+ ions in the commercial production of aluminum?
Ans: H+ ions in the aqueous solution will be discharged at the cathode in preference to the Al3+ ions OR the reduction potential of H+ ions is higher than that of Al3+.
- Calculate the pH of 0.100 moldm-3 solution of a sodium salt of an alkanoic acid whose pKa is 4.80.
Ans: pKb of the conjugate base = 14.0 – 4.80 = 9.20
pOH = ½ pKb – ½ log Cb (Cb = 0.100 moldm-3)
= 9.20/2 + 0.500 = 5.10
Hence pH = 14.0 – 5.10 = 8.90.
Preamble to all schools:
In the commercial preparation of trioxonitrate(V) acid, nitrogen(IV) oxide dissolves in water to give the acid while nitrogen(II) oxide is evolved. The balanced equation for the reaction is as follows:
3NO2 + H2O → 2HNO3 + NO.
O = 16.0; N = 14.0; H = 1.00. The molar volume of a gas at STP is 22.4dm3.
- Calculate the mass of HNO3 that can be produced from 69.0g of NO2
Ans: 3NO2 + H2O → 2HNO3 + NO.
3 x 46.0 2 x 63.0
138g of NO2 ≡ 126g of HNO3
Therefore, mass of HNO3 to be produced from 69.0g of NO2 = (69.0/138)*126g = 63.0g
- Find the volume of NO2 at STP that can be used to produce 189g of HNO3.
Ans: 3NO2 + H2O → 2HNO3 + NO.
Moles of HNO3 in 189g of it. = 189/63.0 = 3.00
Moles of NO2 needed = (3.00/2.00)*3 = 4.50.
Hence volume of NO2 at STP required = 4.50*22.4 = 100.8dm3 = 101dm3
- Calculate the mass of the minimum amount of water required to react completely with 44.8 dm3 of NO2 at STP.
Ans: 3NO2 + H2O → 2HNO3 + NO.
Moles of NO2 in 44.8dm3 of the gas at STP = 44.8/22.4 = 2.00
Moles of water required = (1/3)*2 = 2/3
Mass of water = (2/3)*18.0 = 12.0g
- What is the name of a saturated hydrocarbon with 12 carbons in a chain?
Ans: Dodecane.
- Calculate the number of moles of limestone (CaCO3) in a 2.00 kg gold ore that contains only 0.500% gold, the rest being CaCO3 in the form of limestone.
Ca = 40.0; O = 16.0; C = 12.0
Ans: Mass of limestone = (2000 – 10.0) = 1990
Formula mass of limestone = 40.0 + 12.0 + 48.0 = 100
Moles of limestone = 1990/100 = 19.9 mol.
- Which flask is needed to perform suction filtration in the laboratory?
Ans: (Heavy wall) filtering flask.
Preamble to all schools
Consider the following reversible chemical reaction:
A(g) + 2B(g) 3C(g); ΔHr = -230kJ
- How will an increase in temperature affect the equilibrium constant?
Ans: Keq will decrease since the forward reaction is exothermic.
- At equilibrium how will an increase in volume affect the equilibrium position?
Ans: Since the number of gaseous reactant molecules are the same as the products the equilibrium position will not be affected by changes in volume.
- How will an addition of a catalyst affect the equilibrium constant?
Ans: The equilibrium constant will not be affected by the addition of a catalyst.
Preamble to all schools:
An organic compound A decomposes by second order kinetics. The integrated form of the second order rate equation is 1/[At] = kt + 1/[A0], where [A0]is the initial concentration of A and [At]is its concentration at time t.
- Calculate the rate constant if 10.0% of the initial concentration of 0.100 moldm-3 of A is decomposed after 50.0 seconds. Remember to give your answer in the standard or scientific form.
Ans: 1/[At] = kt + 1/[A0]
Concentration of A after 50.0 s = 0.0900moldm-3.
1/0.0900 = k*50.0 + 1/0.100; 11.1 = 50.0k + 10.0
k = 1.10/50.0 = 1.10/50.0 = 2.20*10-2 mol-1dm3s-1
- Calculate the rate constant if in another reaction 60.0% of the 0.100 moldm-3 of A had decomposed in 150 seconds. Give your answer in the standard form.
Ans: 1/[At] = kt + 1/[A0]
1/0.0400 = k*150 + 1/0.100; 25.0 = 150k + 10.0
k = 15.0/150 = 1.00*10-1 mol-1dm3s-1
- For another decomposition reaction of A, determine the time in seconds it takes 50.0% of the initial concentration of 0.100 moldm-3 of A to decompose if the rate constant of that reaction is found to be 2.00*10-2 mol-1dm3s-1
Ans: 1/[At] = kt + 1/[A0]
1/0.0500 = k*t + 1/0.100; 20.0 = 2.00*10-2t + 10
t = 10.0/2.00*10-2 = 500s (or 8min 20s)
[Alternative method: Time for 50% of concentration to be consumed is t½.
But for a second order reaction, t½ = 1/k[A0]
t½ = 1/2.00*10-2*0.100 = 1.00*103/2 = 500s ]
NSMQ PAST QUESTION; CHEMISTRY (02)
NSMQ Past Question; Chemistry (03)
NSMQ Past Question; Chemistry (01)
Join us on Our WhatsApp platform for more
Biology, Mathematics and Physics will be loaded soon.
About Author
Hi there very cool web site!! Man .. Beautiful .. Superb .. I will bookmark your site and take the feeds also? I am glad to search out so many helpful information here within the post, we want develop extra techniques on this regard, thanks for sharing. . . . . .