Microsoft word - session 1 - practice ans.docx
Session 1 - Practice Answers
4 (aq) + 8H+ → 5Fe3+(aq) + Mn2+(aq) + 4H2O(l)
3 (aq) + 6H+ → 3Sn4+(aq) + I−(aq) + 3H2O
d. 8NH3(g) + 6NO2(g) → 7N2(g) + 12H2O(g)
3 (aq) + 6H+ → 5Bi2+(aq) + 3MnO4 (aq) + 3H2O
g. 6KI(aq) + 8HNO3(aq) → 3I2(aq) + 4H2O(l) + 6KNO3(aq) + 2NO(g)
h. 3C2H5OH(aq) + 2K2Cr2O7 + 8H2SO4 → 3CH3COOH(aq) + 2K2SO4(aq) + 2Cr2(SO4)3(aq) + 11H2O(l)
i. C2H5OH + H2O → CH3COOH + 4H+ + 4e−
The ionic half-reaction method is tedious and involves three ionic half-reactions. Instead, coefficients can
be found logically. It requires 5HNO3 to produce 2NO and 3NO2. The difference in oxygen is
5×3 − 2×1 − 3×2 = 7, hence 7H2O, 14HNO3, and finally 9Cu.
9Cu + 14HNO3 → 9Cu(NO3)2 + 2NO + 3NO2 + 7H2O
This is an example of a reaction taking place in alkaline (basic) solution. Therefore, OH−/H2O is used
3 (aq) + 8Al(s) + 5OH−(aq) + 18H2O(l) → 8Al(OH)4 (aq) + 3NH3(g)
The reduction half-equation is trickier. We can form and solve mass balance equations with unknown
In this case using ionic half-reactions to balance is quite difficult. Instead a shorter method is to use
oxidation states of nitrogen: N+5 + 8e− → N−3.
X is an oxide of sulfur, either SO2 or SO3. We know that SO3 does not form directly from S, rather it is
produced from SO2 in the presence of V2O5 catalyst. The oxidation of SO2 to SO3 is much more difficult,
and therefore it is more likely that oxide X is SO2.
There are two options for Y: either SCl2 or SOCl2. The oxidation state of P in the second reaction does
not change, i.e. it remains +5, meaning the reaction is not redox. S is +4 in SO2, thus Y is SOCl2.
SOCl2 then reacts with HF forming SOF2 and HCl. This is analogous to displacement of halogens in
compounds by fluorine. S and O are electron rich, and are more attractive to F.
SO3, P4S6 and PF3
SO2, SCl2 and SF2
SO3, SCl2 and SOF2
SO2, SOCl2 and SOF2
Unfortunately there was a typo: mass of Na2CO3 should have been 10.6 g, not 12.6 g. If you got 0.288
M, then your solution is correct.
This question is a good example of importance of stoichiometry. The very reason of knowing reaction
coefficients is to be able to know the right amounts of substances consumed and produced in reactions.
Dissolution and dissociation of the salts can be written as:
From the gas laws, P
is inversely proportional to V
, or in other terms, PV
= constant, if the amount of gas
and temperature do not change. Here, however, the amount of gas changes. One possible solution is to
treat gases from each bulb separately and calculate the pressure exerted by each gas. Note that the
same gas could be used in both bulbs, but the solution is the same. If bulb 2 is empty, then the new
pressure exerted by the gas in bulb 1 will be:
1.00 L × 1.50 atm = 5.00 L × P
1 ⇒ P
1 = 0.300 atm
4.00 L × 2.00 atm = 5.00 L × P
2 ⇒ P
2 = 1.60 atm
Total pressure, P
1 + P
2 = 1.90 atm
The concept of partial pressures is very important for understanding gas mixtures and diffusion. Partial
pressure of an ideal gas in a gas mixture is the pressure exerted by the gas as if it was by itself in the
container. The total pressure is the sum of partial pressures.
The molecular weight of a gaseous mixture is the sum of molar fractions multiplied by molecular weight
of every component, MWmixture = ∑xiMWi. Therefore: MWair = 0.2×32 + 0.8×28 = 28.8 g/mol.
At rtp (25 °C and 1 atm) 1 mole of gas occupies 24.4 L. 1 mole of air weighs 28.8 g. The density is thus:
Many students learn that 1 mol of gas = 24 L (correct to 2 s.f.). The answer would be 1.20 g/L. This is
Calibration graphs are much used in science and even more so in engineering. The idea is to test many
samples with known parameters to produce a calibration graph. It can then be used to identify unknown
mixture = 103.639 − 94.604 = 9.035 g
From the plot, the mixture with such density has about 60% of propanone by volume. Water is thus about
2 moles of gaseous reagents form 2 moles of gaseous products. This has no effect on pressure.
100 ml of water can dissolve up to 87.5 g of NaNO3 at 20 °C, therefore 20.0 ml of water can only dissolve
17.5 g of NaNO3. Similarly 20 ml of water can dissolve only 15.6 g of LiCl. The initial mixture contained
20.0 g of each salt. The precipitate therefore contains 2.5 g of NaNO3 and 4.4 g of LiCl.
Organic molecules tend to be big and the short-hand used in organic chemistry avoids showing
unimportant hydrogens and carbons. The full structure of methocarbamol:
The condensed formula is C11H14NO5. Even without knowing this, you should be able to count the 5
oxygen atoms and use the molecular weight (molar mass) provided to calculate mass percentage of
Since the number of moles of Ca and CaSO4 are the same (i.e. 1 to 1):
In 1.00 ton of the mineral, the mass of Na2SO4 can be found using simple proportions:
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Journal of Nutritional & Environmental MedicineMay 2007; 16(2): 149–166MARGARET MOSS, MA (CANTAB), UCTD (MANCHESTER), DIPION, CBIOL,MIBIOL, Director of the Nutrition and Allergy Clinic11 Mauldeth Close, Heaton Mersey, Stockport, Cheshire SK4 3NPAbstractPurpose: To collate evidence on nutrient deficiencies caused by drugs. Design: Search of Medline and other databases, and published litera