JAMB UTME - Chemistry - 1985
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A colourless organic compound X was burnt in excess air to give two colourless and odourless gases, Y and Z, as products. X does not decolorize blue bromine vapour, Y turns lime milky while Z gives a blue colour with copper (ll) tetraoxosulphate (IV).
Compound X is?
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The oxidation of
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The diagram shown above represents the solubility curves of two salts, X and Y, in water, use this diagram to answer the question. If 80 g each of X and Y are taken up in 100 g of water at 353 K we shall have
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Hydrogen 40,000
Helium 3,100
Oxygen 22
Neon 8.6
Nitrogen 6.6
Carbon 3.5
Silicon 1
Magnesium 0.91
Iron 0.6
Sulfur 0.38
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Use the following option above to answer this question. Which of the following is the functional group of carboxylic acids?
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Copper sulphate solution is electolysed using platinum electrodes A. current of 0.193 amperes is passed for 2hrs. How many grams of copper are deposited? (cu = 63.5, F = 96500 coulombs)
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current = 0.193amps
Quantity of electricity = 120 x 0.193
= 60 x 23.16
96,500 deposits 63.5 gms of Cu
60 x 23.2 will deposit 63.5×60×23.296500
= 0.457
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In the reaction \( \mathrm{H_2(g)O \rightarrow H_2(g) + \frac{1}{2}O_2(g)} \)
\( \Delta H = -2436000\ \mathrm{KJ} \), which of the following has no effect on the equilibrium position?
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The I.U.P.A.C name for the compound
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A colourless organic compound X was burnt in excess air to give two colourless and odourless gases, Y and Z, as products. X does not decolorize blue bromine vapour, Y turns lime milky while Z gives a blue colour with copper (ll) tetraoxosulphate (IV).
Y and Z are respectively?
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1000gm will contain 1000160 = 6.25 moles
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A certain industrial process is represented by the chemical equation \(2\mathrm{A}(g) + \mathrm{B}(g) \rightarrow \mathrm{C}(g) + 3\mathrm{D}(g)\quad \Delta H = X\ \mathrm{kJmol}^{-1}\). Which of the following conditions will favour the yield of the products?
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Dividing by the atomic weight we have for C = 51 = 6
H = 121 = 12
O = 1616 = 1
Divide by the smallest number, we have = 61 = 6
H = 121 = 12
O = 11 = 1
The formula is C6H12O
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i.e P1V1298=4P1×4V1T2
T2 = 298×4P1×4V1P1V1
= 298 + 16 = 4768.0k
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The diagram shown above represents the solubility curves of two salts, X and Y, in water, use this diagram to answer the question. At room temperature (300 K)
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Use the above options to the question. The products formed on hydrolysis of
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The diagram shown above represents the solubility curves of two salts, X and Y, in water, use this diagram to answer the question. If the molar mass of X is 36 g, the number of moles of X dissolved at 343 K is
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Figure 1 above shows part of the periodic Table which of the elements belongs to the p-block?
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