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oee [108]
3 years ago
6

Complete the equation for the ionization of water by drawing the conjugate acid and conjugate base. Include lone pairs of electr

ons. Two water molecules react to form the conjugate acid which has a plus 1 charge, and the conjugate base, which has a minus one charge. Each water molecule consists of a central oxygen atom bonded to two hydrogen atoms. There are two lone pairs on the oxygen atom.

Chemistry
2 answers:
kodGreya [7K]3 years ago
5 0

Answer:

Here's what I get  

Explanation:

I followed the instructions and got the diagram below.

nydimaria [60]3 years ago
3 0

Please find detailed answers in the image below

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Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of
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Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

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Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

dw=Pdv

The pressure is constant, so:  w=P(v_{2}-v_{1} )=0.5*100*1.5=75\frac{kJ}{kg} (There is a multiplication by 100 due to the conversion of bar to kPa)

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On the other hand, due to the low pressure the ideal gas law may be appropriate. The ideal gas law is written for both states:

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}\\V_{2}=2.5V_{1}\\P_{2}=P_{1}\\2.5P_{1}V_{1}=nRT_{2}  

Subtracting the first from the second:

1.5P_{1}V_{1}=nR(T_{2}-T_{1})

Isolating T_{2}:

T_{2}=T_{1}+\frac{1.5P_{1}V_{1}}{nR}

Assuming that it is water steam, n=0.1666 kmol

V_{1}=\frac{nRT_{1}}{P_{1}}=\frac{8.314*0.1666*373.15}{500} =1.034m^{3}

T_{2}=100+\frac{1.5*500*1.034}{0.1666*8.314}=659.76 ºC

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