1)The cubic form for the fictitious element Smeltium is FCC, the atomic radius i
ID: 881973 • Letter: 1
Question
1)The cubic form for the fictitious element Smeltium is FCC, the atomic radius is 141.8 pm and the molar mass is 255.7 g/mol. Determine the density of Smeltium.
2)
A student combines KI (aq) and Pb(NO3)2 (aq). What is the NIE? What are the spectator ions?
3)
What is the rate expression for the combustion of propane?
4)
D + E The following are initial rate data for: ExperimentInitial [AInitial B 3A + 2B + C Initial Cl Initial Rate 0.10 0.10 0.20 0.10 The rate law is Rate = k[ANB]"[CP 0.10 0.20 0.10 0.10 0.10 0.10 0.10 0.20 2.31 9.24 4.62 2.31 2 3 4Explanation / Answer
1) fictitious element Smeltium is FCC, the atomic radius is 141.8 pm and the molar mass is 255.7 g/mol. Determine the density of Smeltium.
Answer : The number of atoms per unit cell for FCC = 4 = Z
We know that
density = Z X M / Na X a3
a = edge length = radius 2(2)1/2
Na = 6.023 X10^23
so density = 4 X 255.7 / (141.8 X 10-12 )3 (2.828)3 X 6.023 X1023
Density = 1022.8 / 2.856 X 10^6 X 10^-36 X 22.617 X 6.023 X 1023
Density = 2.628 X 10^6 g / m3
Density = 2.628 g / cm3
2) A student combines KI (aq) and Pb(NO3)2 (aq). What is the NIE? What are the spectator ions?
the equation will be KI (aq) + Pb(NO3)2 (aq) --> 2KNO3 (aq) + PbI2(s)
NIE :
Pb+2 (aq) + 2I- (aq)--> PbI2(s)
Spectator ions = K+ and NO3-
3) What is the rate expression for the combustion of propane
C3H8 + 5O2 --> 3CO2 + 4H2O
4) We can determine the order of reaction be considering the effect of change of concentration of both the reactants if the concentration of other is kept constant
1) [A] = 0.10 M [B] = 0.10 [C] =0.10 M Rate = 2.31
[A] = 0.20 M [B] = 0.10 [C] =0.10 M Rate = 9.24
the concentration of B and C is kept constant and doubling the concentration of A , the rate becomes = 4 times
so order of reaction with respect of A is 2
2) [A] = 0.10 M [B] = 0.10 [C] =0.10 M Rate = 2.31
[A] = 0.10 M [B] = 0.20 [C] =0.10 M Rate = 4.62
the concentration of A and C is kept constant and doubling the concentration of B , the rate becomes = 2 times
so order of reaction with respect of B is 1
3) [A] = 0.10 M [B] = 0.10 [C] =0.10 M Rate = 2.31
[A] = 0.10 M [B] = 0.10 [C] =0.20 M Rate = 2.31
the concentration of A and B is kept constant and doubling the concentration of C , the rate becomes same
so order of reaction with respect of B is 0
So overall rate law will be:
Rate = K [A]2[B][C]0
2.31 = K [0.1]2[0.1]
K = rate constant = 2310
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