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Question 1 Report
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Question 3 Report
In a reaction: SnO + 2C → Sn + 2CO, the mass of coke containing 80% carbon required to reduce 0.302 kg of pure tin oxide is?
(Sn - 119, O - 16, C - 12).
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Mass of coke is required not mass of carbon,so it is (0.048× 100)÷ 80 = 0.06
Question 4 Report
Select the right answer from the structure above.Which of the following compounds represents the polymerization product of ethyne?
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Question 6 Report
The solubility in mol dm-3 of 20g of CuSO4 dissolved in 100g of water at 180o is
[Cu = 64, S = 32, O =16]
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Question 9 Report
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P1 = 0.971 atmosphere
V2 = ? T2 = 273K P2 = 1 atom
V2 = (0.971 atom χ 146cm3 χ 273k)/(1atm χ 291k) = 132997 0r = 133cm3
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Question 17 Report
0.1 faraday of electricity deposited 2.95 g of nickel during electrolysis of an aqueous solution.
Calculate the number of moles of nickel that will be deposited by 0.4 faraday? (Ni = 58.7)
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Question 19 Report
Use the graph above to answer this question. The structure of cis-2-butene is
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Question 20 Report
Use the above graph above to answer this question . \(\mathrm{CH_3-C=CH}\ \xrightarrow{\frac{\mathrm{Na}}{\mathrm{liqNH_3}}}\ \mathrm{P}\), Compound P, in the above reaction, is
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Question 23 Report
In the reaction:
2Hl(g) ↔ H2(g) + l2(g)' ΔH = 10 kj;
The concentration of iodine in the equilibrium mixture can be increased by?
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Question 24 Report
\[ \mathrm{Fe_2O_{3(s)} + 2Al_{(s)} \rightarrow Al_2O_3 + 2Fe_{(s)}} \]
If the heats of formation of \( \mathrm{Al_2O_3} \) and \( \mathrm{Fe_2O_3} \) are - 1670 kj mol\(^{-1}\) and - 822 kj mol\(^{-1}\) respectively, the enthalpy change in kj for the reaction is ?
Question 25 Report
.....................P....Q.....R......S
Proton..............13....16....17.....19
Electron............13....16....17.....19
Neutron.............14....16....35.....20Which of the four atoms P,q,R and in S in the above data can be described by the following properties relative atomic mass is greater than 30 but less than 40; it has an odd atomic number and forms a unipositive ion in solution?
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Question 27 Report
10.0 dm3 of air containing H2S as an impurity was passed through a solution of Pb(NO3)2 until all the H2S had reacted. The precipitate of PbS was found to weigh 5.02 g. According to the equation: Pb(NO3)2 + H2S → PbS + 2HNO3 the percentage by volume of hydrogen sulphide in the air is?
(Pb = 207, S = 32, GMV at s.t.p = 22.4dm3)
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Question 30 Report
Changes in the physical in the scheme above.The letter X,Y and Z respectively represent
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Question 32 Report
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32g of O2 = mole of O2
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Question 36 Report
Equal volumes of CO,SO2, and H2S were released into a room at the same point and time. Which of the following gives the order of diffusion of the gases to the opposite corner of the room?
(S = 32, C = 12, O = 16, N = 14, H = 1)
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Question 42 Report
Chlorine, consisting of two isotopes of mass numbers 35 and 37, has an atomic mass of 35.5.
The relative abundance of the isotope of mass number 37 is?
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Question 43 Report
What is the IUPAC name for the hydrocarbon?
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Question 49 Report
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