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Question 1 Rapport
R ∝ 1√Mm or R ∝ 1√D
Dx = 0.5gdm-3, Dy = 2gdm-3
R= K√D
R√D = k
R1√D1 = R1√D2
Rx√Dx = Ry√Dy
RxRy = √Dy√Dx
= √2√0.5
= 2.0
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Question 13 Rapport
CO2 combines with water to give a weak trioxocarbonate (IV) acid.
CO2 + H2O ? H2CO3
Question 14 Rapport
This deals with a two forms of a molecule where the chemical connectivity is the same but the electrons are distributed differently around the structure.
Question 15 Rapport
Question 16 Rapport
k = concentration of productconcentration of reactant
Question 17 Rapport
Mm of XO2 = x + 16(2) = x + 32
vapour density = 32
∴ x + 32 = 32 × 2
x + 32 = 64
x = 64 - 32
x = 32
∴ the relative molecular mass of X is 32
Relative molecular mass = vapour density × 2
Question 18 Rapport
Question 19 Rapport
\( {}^{226}_{88}\mathrm{Ra} \rightarrow {}^{x}_{86}\mathrm{Rn} \) + alpha particle
Question 20 Rapport
Question 21 Rapport
It contains all the functional group of protein which includes the amino, alkanol and the carboxylic group. Adding few drops of conc HNO3 to a protein, gives an intense yellow colouration.
It is called Xanthopreitic test.
Question 22 Rapport
X(g) + 3Y(g) ---- 2z(g) H = +ve. if the reaction above takes place at room temperature, the G will be
Enthalpy Change [?H]Entropy Change [?S]Gibbs free Energy[?G]PositivePositivedepends on T, may be + or -NegativePositivealways negativeNegativeNegativedepends on T, may be + or -PositiveNegativealways positive
?G= ?H ? T?S.
To determine whether ?G will be positive or negative, the value of ?H(change in enthalpy) and ?S (change in entropy) must be given. Likewise the temperature.
Question 23 Rapport
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Question 29 Rapport
Amount in moles = n
Volume v = 10.5dm3
Pressure P = 6atm
Temperature T = 30°C + 273 = 303k
R, Gas constant = 0.082 atmdm3k-1 mol
Recall from ideal gas equation
pv = nRT
n = RVRT
n = 6×1050.082×303
n= 2.536mol
Question 30 Rapport
Question 31 Rapport
In the diagram above. X is
Question 32 Rapport
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Question 37 Rapport
= MmIT96500n
Copper II Chloride = CuCl2
CuCl2 → Cu2+ + 2Cl2
Mass of compound deposited = Molar mass × Quantity of Electricity96500 × no of charge
Q = IT
I = 0.5A
T = 45 × 60
T = 2700s
Q = 0.5 × 2700
= 1350c
Molarmass = 64gmol-1
no of charge = + 2
Mass = 64×135096500×2
Mass = 0.448g
Question 38 Rapport
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