1. What trends do you see when considering total electrons, valence electrons (highest occupied energy level/outermost level ele
2 answers:
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Answer:
a) 16.2 dm^3/mol*h
b) 6.1 × 10^3 s, 2.5 × 10^3 s (it is different to the hint)
Explanation:
We can use the integrated rate equation in order to obtain k.
For the reaction A+ B --> P the reaction rate is written as
Rate =
If and
are the inital concentrations and x the concentration reacted at time t, so
and
and the rate at time t is written as:
Separating variables and integrating
The integral in left side is solved by partial fractions, it can be used integral tables
Using logarithm properties (ln x - ln y = ln(x/y))
Using the given values k can be calculated. But the data seems inconsistent since if the concentration of A changes from 0.075 to 0.02 mol dm^-3 it implies that 0.055 mol dm^-3 of A have reacted after 1 h, so according to the reaction given the same quantity of B should react, and we only have a of 0.05 mol dm^-3.
Assuming that the concentration of B fall to 0.02 mol dm^-3 (and not the concentration the A). So we arrive to the answer given in the Hint.
So, the values given are t= 1, ,
,
, it implies that the quantity reacted, x, is 0.03 and
. Then, the value of k would be
b) the question b requires calculate the time when the concentration of the specie is half of the initial concentration.
For reactant A, It is solved with the same equation
but suppossing that so
, k=16.2 and the same initial concentrations. Replacing in the equation
For reactant B, so
, k=16.2 and the same initial concentrations. Replacing in the equation
Note: The procedure presented is correct, despite of the answer be something different to the given in the hint, I obtain that result if the k is 19.2... (maybe an error in calculation of given numbers)
Explanation:
For an isothermal process equation will be as follows.
W = nRT ln
It is given that mass is 10 kg/s or 10,000 g/s (as 1 kg = 1000 g). So, calculate number of moles of water as follows.
No. of moles =
=
= 555.55 mol/s
= 556 mol/s (approx)
As T = or (50 + 273.15) K = 323.15 K. Hence, putting the given values into the above formula as follows.
W = nRT ln[/tex]\frac{P_{1}}{P_{2}}[/tex]
=
=
= -3440193.809 J/s
Negative sign shows work is done by the pump. Since, 1 J = 0.001 kJ. Therefore, converting the calculated value into kJ as follows.
= 3440.193 kJ/s
= 3451 kJ/s (approx)
Thus, we can conclude that the pump work is 3451 kJ/s.