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Ibeere 1 Ìròyìn
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
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Ibeere 9 Ìròyìn
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.
Ibeere 10 Ìròyìn
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.
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Ibeere 19 Ìròyìn
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.
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Ibeere 23 Ìròyìn
In the diagram above, Y is
Ibeere 24 Ìròyìn
The IUPAC nomenclature of the structure above is
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Ibeere 28 Ìròyìn
\( {}^{226}_{88}\mathrm{Ra} \rightarrow {}^{x}_{86}\mathrm{Rn} \) + alpha particle
Ibeere 29 Ìròyìn
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
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Ibeere 31 Ìròyìn
The constituent of air includes O2, CO2, N2 and Noble gases and for rusting process to take place.
The presence of O2 and CO2 as a constituent of air is indispensable
Ibeere 32 Ìròyìn
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
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Ibeere 36 Ìròyìn
In the diagram above. X is
Ibeere 37 Ìròyìn
= 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
Ibeere 38 Ìròyìn
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
Ibeere 39 Ìròyìn
k = concentration of productconcentration of reactant
Ibeere 40 Ìròyìn
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