JAMB UTME - Chemistry - 1997
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∆G = 89 - 298 x 11.8 = -3427.4
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2(-297) 2(-396)
-594 -792
Heat change = heat of product - reactants = -792 - (-594) = 198kj mol-
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A sample of a substance containing only C and H burns in excess O2 to yield 4.4g of CO2
And 2.7 g of H2O. The empirical formula of a substance is?(C = 12, O = 16, H = 1)
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An element, X, forms a volatile hydride XH3 with a vapour density of 17.0. The relative atomic mass of X is?
(H = 1)
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∴ Relative Atomic Mass of X in XH3 = 34 - 3
= 31.0 ∴ x = 31.0
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Given that 15.00 cm3 of H2
SO4 was required to completely neutralize 25.00 cm3 of 0.125 mol dm-3 NaOH, calculate the molar concentration of the acid solution:
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(MaVa)/(MBVB) (1)/(2) (MA x 15)/(0.125 x25) = 1/2
MA = 0.104
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The IUPAC nomenclature of the organic compound with the above structural formula is
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Choose the correct answer from the options above.
When Fehling's solution is added to two isomeric carbonyl compounds X and Y with the molecular formula C 5H 10O, compound X gives a red precipitate while Y does not react. It can be inferred that X is
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The compound above contains
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Use the graph above to answer this question.
A sample, X, solid at room temperature, was melted, heated to a temperature of 358 K and allowed to cool as shown in OPQR. The section OP suggests that X is in the
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\(2\mathrm{H_2S(g)} + \mathrm{SO_2(g)} + \mathrm{H_2O(l)} \rightarrow 3\mathrm{S(s)} + 3\mathrm{H_2O(l)}\)..........(I)
\(3\mathrm{CuO(s)} + 2\mathrm{NH_3(g)} \rightarrow 3\mathrm{Cu(s)} + 3\mathrm{H_2O(l)} + \mathrm{N_2(g)}\).........(II)
In the equations above, the oxidizing agent (I) and the reducing agent in (II) respectively are?
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How many more isomers of the compound above can be obtained ?
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120g (MgSo4) contains - 18g (water)
20g contain (41 - 20g) water.
120g → 18 y; 20 → 21
∴ 18 y = (126)/(18) = 7
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25cm3 require 0.03 g of CnH2n + 1 COOH 1000CM3 → x x = (1000) / (25) x 0.03 = 1.2g
CnH2n + 1COOH + KOH → CnH2n + 1 COOK + H2O
from (MAVA) / (MB x VB) = (1) / (1) (MA x 1.2) / (0.02 x 0.78) = (1) / (1)
MA = 0.013
Con. = Molarity x Molecular Mass
Molecular Mass = (0.78) / (0.013) x 60
CnH2n + 1 COOH = 60
12n + 1 (2n + 1) + 12 + 32 + 1 = 6
14n = 60 - 46; n = (14) / (14) = 1
∴ CH3COOH
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Use the graph above to answer this question.
A sample, X, solid at room temperature, was melted, heated to a temperature of 358 K and allowed to cool as shown in OPQR. The section PQ indicates that X is
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200 cm3 of air was passed over heated copper in a syringe several times to produce copper (II) oxide.
When cooled, the final volume of air recorded was 158cm3. Estimate the percentage of oxygen in the air?
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(60/t) (30/7)/ √(35.5)/(16) t = (60/30)/(7) x (4)√35.5
t = (60 x 7 x √35.5)/(30 x 4) = 21s
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2X(aq) + MnO2(s) + 4H+(aq) → X2(g) + Mn2+(aq) + 2H2O.
The reaction above can be used for the laboratory preparation of all halogens except fluorine because it is?
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