¿que ocurre con el electrón de Valencia del K+ en el compuesto bromuro de potasio KBr
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This problem is describing a gas mixture whose mole fraction of hexane in nitrogen is 0.58 and which is being fed to a condenser at 75 °C and 3.0 atm, obtaining a product at 3.0 atm and 20 °C, so that the removed heat from the system is required.
In this case, it is recommended to write the enthalpy for each substance as follows:
Whereas the specific heat of liquid and gaseous n-hexane are about 200 J/(mol*K) and 160 J/(mol*K) respectively, its condensation enthalpy is 31.5 kJ/mol, boiling point is 69 °C and the specific heat of gaseous nitrogen is about 29.1 J/(mol*K) according to the NIST data tables and and
are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:
It is seen that the heat released by the nitrogen is neglectable in comparison to n-hexanes, however, a rigorous calculation is being presented. Then, we add the previously calculated enthalpies to compute the amount of heat that is removed by the condenser:
Finally we convert this result to kJ:
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Answer:
see explanation below
Explanation:
Question is incomplete, so in picture 1, you have a sample of this question with the missing data.
Now, in general terms, the absorbance of a substance can be calculated using the beer's law which is the following:
A = εlc
Where:
ε: molar absortivity
l: distance of the light in solution
c: concentration of solution
However, in this case, we have a plot line and a equation for this plot, so all we have to do is replace the given data into the equation and solve for x, which is the concentration.
the equation according to the plot is:
A = 15200c - 0.018
So solving for C for an absorbance of 0.25 is:
0.25 = 15200c - 0.018
0.25 + 0.018 = 15200c
0.268 = 15200c
c = 0.268/15200
c = 1.76x10⁻⁵ M
