Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,00
0 and 100,000 g/mola indicated below: (A) Equal numbers of molecules of each sample (B) Equal masses of each sample. For each of the mixtures, calculate the number-average and weight-average molar masses and comment upon the meaning of the values.
1 answer:
Answer:
Hello some parts of your question is missing
Two mixtures were prepared from three very narrow molar mass distribution polystyrene samples with molar masses of 10,000, 30,000 and 100,000 g mol−1 as indicated below: (a) Equal numbers of molecules of each sample (b) Equal masses of each sample (c) By mixing in the mass ratio 0.145:0.855 the two samples with molar masses of 10,000 and 100,000 g mol−1 For each of the mixtures, calculate the number-average and weight-average molar masses and comment upon the meaning of the values.
Answers:
a) 46.666.66 g/mol
b) 20930.23 g/mol
c)43333.33 g/mol
Explanation:
A)The equal number of molecules of each sample can be calculated using
because for the number of molecules to be equal : n1 = n2 = n3 = n
Mn = 46666.66 g/mol
B ) To calculate the equal masses of each sample
we apply this equation
ATTACHED BELOW IS THE REMAINING PART OF THE SOLUTION
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The question is incomplete. The complete question is :
A common "rule of thumb" for many reactions around room temperature is that the rate will double for each ten degree increase in temperature. Does the reaction you have studied seem to obey this rule? (Hint: Use your activation energy to calculate the ratio of rate constants at 300 and 310 Kelvin.)
Solutions :
If we consider the activation energy to be constant for the increase in 10 K temperature. (i.e. 300 K → 310 K), then the rate of the reaction will increase. This happens because of the change in the rate constant that leads to the change in overall rate of reaction.
Let's take :
The rate constant = respectively.
The activation energy and the Arhenius factor is same.
So by the arhenius equation,
and
Given, J/mol
R = 8.314 J/mol/K
∴
So, no this reaction does not seem to follow the thumb rule as its activation energy is very low.
Answer:
Titration
Explanation:
The best technique which can be used to determine the number of moles of the HCl in the sample is titration.
The given amount of HCl solution must be titrated with known concentration of the base like NaOH.
The volume of NaOH required must be noted also.
According to the reaction,
At equivalence point
Moles of = Moles of
Considering:-
Moles of
Thus, in this way, moles of HCl can be determined.