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Ajụjụ 1 Ripọtì
= 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
Ajụjụ 2 Ripọtì
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.
Ajụjụ 3 Ripọtì
Ajụjụ 4 Ripọtì
Ajụjụ 5 Ripọtì
Ajụjụ 6 Ripọtì
The IUPAC nomenclature of the structure above is
Ajụjụ 7 Ripọtì
CO2 combines with water to give a weak trioxocarbonate (IV) acid.
CO2 + H2O ? H2CO3
Ajụjụ 8 Ripọtì
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Ajụjụ 14 Ripọtì
Ajụjụ 15 Ripọtì
CH2S(g) + O2(g) → 2Cn + SO2(g)
What is the change in the oxidation number of copper in the reaction?
Cu2S
2 Cu - 2(1) = 0
2 Cu = 2
Cu = 22
Cu = +1
In the product, Cu
Cu = O
The oxidation number of Cu in Cu2S and Cu respectively is +1 and 0 respectively
Ajụjụ 16 Ripọtì
\( {}^{226}_{88}\mathrm{Ra} \rightarrow {}^{x}_{86}\mathrm{Rn} \) + alpha particle
Ajụjụ 17 Ripọtì
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Ajụjụ 21 Ripọtì
Ajụjụ 22 Ripọtì
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.
Ajụjụ 23 Ripọtì
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Ajụjụ 26 Ripọtì
Ajụjụ 27 Ripọtì
Ajụjụ 28 Ripọtì
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.
Ajụjụ 29 Ripọtì
k = concentration of productconcentration of reactant
Ajụjụ 30 Ripọtì
Ajụjụ 31 Ripọtì
Ajụjụ 32 Ripọtì
Ajụjụ 33 Ripọtì
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
Ajụjụ 34 Ripọtì
Ajụjụ 35 Ripọtì
Ajụjụ 36 Ripọtì
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
Ajụjụ 37 Ripọtì
In the diagram above. X is
Ajụjụ 38 Ripọtì
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
Ajụjụ 39 Ripọtì
Ajụjụ 40 Ripọtì
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