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ziro4ka [17]
3 years ago
7

Based on the types of elements in the compound are there ionic or covalent? A. KCI B.NO2

Chemistry
1 answer:
Alex3 years ago
7 0

B.

Explanation:

iehebrkee keen enensjsb sh sry need to get the points?

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A gas stream containing n-hexane in nitrogen with a relative saturation of 0.58 (as a fraction, multiply by 100% if you prefer %
kondor19780726 [428]

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:

H_{C-6}=y_{C-6}C_v(T_b-Ti)+\Delta _vH+C_v(T_f-Tb)\\\\H_{N_2}=y_{N_2}C_v(T_f-Ti)

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 y_{C-6} and y_{N_2} are the mole fractions in the gaseous mixture. Next, we proceed to the calculation of both heat terms as shown below:

H_{C-6}=0.58*200(69-75)+(-31500)+160(20-69)=-40036J/mol\\\\H_{N_2}=0.42*29.1(20-75)=-672.21J/mol

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:

Q=-40036+(-672.21)=-40708.21J

Finally we convert this result to kJ:

Q=-40708.21J*\frac{1kJ}{1000J}\\\\Q=-40.7kJ

Learn more:

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2 years ago
How many grams of water will form if 10.54 g H2 react with 95.10 g O2?
sergiy2304 [10]

Answer:

94.0 g H2O

Explanation:

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3 years ago
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5 0
3 years ago
Reaction is
-Dominant- [34]
According  to the task, you  are proveded with patial pressure of CO2 and graphite, and here is complete solution for the task :
At first you have to find n1 =moles of CO2 and n2 which are moles of C 
<span>The you go :
</span>CO2(g) + C(s) \ \textless \ =\ \textgreater \  2CO(g)

n1 n2 0
-x -x +2x
n1-x n2-x 2x Kp=P^2 (CO)/Pco2
After that you have to use the formula PV=nRT
 Kp=(2x)^2/n1-x
Then you have to solve x, and for that you have to use <span>RT/V
And to find total values:</span>P(total)=n(total)*R*T/V
I am absolutely sure that this would be helpful for you.
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