FORM FOUR CHEMISTRY NECTA MULTIPLE CHOICE 2013

 THE UNITED REPUBLIC OF TANZANIA NATIONAL EXAMINATIONS COUNCIL

CERTIFICATE OF SECONDARY EDUCATION EXAMINATION 

032/1  CHEMISTRY 1

 (For Both School and Private Candidates) 

Time: 3 Hours  Thursday, 07th November 2013 p.m.

 Instructions

  1. This paper consists of sections A, B and C.
  2.  Answer all questions in this paper.
  3.  Calculators and cellular phones are not allowed in the examination room.
  4.  Write your Examination Number on every page of your answer booklet(s).
  5.  The following constants may be used.

 Atomic masses:

 H = 1, C = 12, O = 16, N = 14, Na = 23, Mg = 24, Al = 26, S = 32, Cl = 35.5, Ca = 40, Fe = 56, Cu = 64, Ag = 108.

 Avogadro’s number = 6.02 x 1023. GMV at s.t.p. = 22.4 dm3.

 1 faraday = 96,500 coulombs.

 Standard pressure = 760 mm Hg.

 Standard temperature = 273 K.

 1 litre = 1 dm3 = 1000 cm3.

SECTION A (20 Marks)

Answer all questions in this section. 

1.  For each of the items (i) – (x), choose the correct answer from the given alternatives and write its letter beside the item number in the answer booklet provided.

 (i)  Which action should be taken immediately after concentrated sulphuric acid spilled on the skin?

  1. Its should be rinsed off with large quantities of running water.
  2. It should be neutralized with solid CaCO3
  3. It should be neutralized with concentrated NaOH.
  4. The affected area should be wrapped tightly and shown to a medical health provider.
  5. It should be neutralized with concentrated KOH.
Choose Answer :

 (ii)  In the titration of a monoprotic acid with a solution of sodium hydroxide of known concentration, what quantities will be equal at the equivalence point?

  1. concentration of hydroxide solution and hydronium ions.
  2. number of moles of hydroxide ions added and number of moles of hydronium ion initially present.
  3. number of moles of hydroxide solution added and volume of acid solution initially present.
  4. number of moles of hydroxide ion added and the number of moles of monoprotic acid initially present.
  5. volume of sodium hydroxide solution added and volume of acid solution initially present.
Choose Answer :

 (iii)  The charge of one mole of electrons is represented by the term

  1.   one ampere
  2. one coulomb
  3. one volt
  4. one faraday E
  5. one gram.
Choose Answer :

 (iv)  65.25 g sample of CuSO4.5H2O (M = 249.7) was dissolved in water to make 0.800 L of solution. What volume of this solution must be diluted with water to make 1.00 L of 0.100 M CuSO4?

  1.   3.27 ml
  2. 383 ml
  3. 209 ml
  4. 65.25 ml
  5. 306 ml.
Choose Answer :

 (v)  Consider the system at equilibrium: H2O(l) H2O(g) for which Which change(s) will increase the yield of H2O(g).

  1. Increase in temperature
  2. Increase in the volume of the container
  3. Increase in temperature and volume of the container
  4. Increasing surface area of oxygen
  5. Increasing surface area of reactants.
Choose Answer :

 (vi)  As water is added to an acid, the acid becomes

  1.   more acidic and its pH goes down
  2. more acidic and its pH goes up
  3. less acidic and its pH goes up
  4. less acidic and its pH goes down
  5. neutral and its pH becomes 7.
Choose Answer :

 (vii)  Three elements, X, Y and Z, are in the same period of the periodic table. The oxide of X is amphoteric, the oxide of Y is basic and the oxide of Z is acidic. Which of the following shows the elements arranged in order of increasing atomic number?

  1.   X, Y, Z
  2. Y, Z, Y
  3. Z, X, Y
  4. Y, X, Z
  5. X, Z, Y.
Choose Answer :

 (viii)  Which of the following compounds contains only two elements?

  1.   Magnesium hydroxide
  2. Magnesium nitride
  3. Magnesium phosphate
  4. Magnesium sulphite
  5. Magnesium sulphate.
Choose Answer :

 (ix)  An atom has 26 protons, 26 electrons and 30 neutrons. The atom has

  1.   atomic number 26, mass number 52
  2. atomic number 56, mass number 30
  3. atomic number 30, mass number 82
  4. atomic number 52, mass number 56
  5. atomic number 26, mass number 56.
Choose Answer :

 (x)  The following equation is a propagation step in the chlorination of methane:

  1. Cl2 Cl + Cl
  2. CH3 + Cl CH3Cl
  3. CH3 + Cl2 CH3Cl + Cl
  4. CH4 + Cl CH3Cl + H
  5. CH3 + Cl2 Ch2Cl + Cl.
Choose Answer :

For Call,Sms&WhatsApp: 255769929722 / 255754805256

   Click Here To Access You Scheme(ONLY IF YOU A HAVE CODE)