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adoni [48]
3 years ago
7

If you change the 2 in front of 2O2 to a 3, what will be the change in the results on the right side of the equation? (1 point)

Chemistry
1 answer:
Simora [160]3 years ago
3 0

Answer:

There is an extra O2 molecule left over

Explanation:

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In a student experiment, a constant-volume gas thermometer is calibrated in liquid nitrogen (−196°c ) and in boiling ethyl alcoh
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The given linear equation is:

P=a+bt

Determine the value of a and b (constants) by plug in the values of pressure and temperature.

P_{1} (Pressure of liquid nitrogen) =0.349 atm

T_{1} (Temperature of liquid nitrogen) =-196^{0}C

P_{2} (Pressure of ethyl alcohol) =1.634 atm

T_{2} (Temperature of ethyl alcohol) =77^{0}C

Put above values in given equation:

P_{1}=a+bT_{1}

0.349 atm=a+b(-196^{0}C)                 (1)

P_{2}=a+bT_{2}

1.634 atm=a+b(77^{0}C)              (2)

Subtract equation (1) from equation (2), we get the value of b

1.285 atm = b(273^{0}C)

b = 4.7\times 10^{-3} atm/^{0}C

Now, put the value of b in equation (1)

0.349 atm=a+(4.7\times 10^{-3}atm/^{0}C)(-196^{0}C)    

a = (0.349) + (4.7\times 10^{-3}atm/^{0}C)(196^{0}C)

a =921.549\times 10^{-3} atm

Now, at absolute zero, Pressure is equal to zero.

Absolute temperature = T_{o} =-\frac{a}{b}

T_{o} =  -\frac{921.549\times 10^{-3} atm}{4.7\times 10^{-3} atm/^{0}C}

= -196.074 ^{0}C

Thus, absolute zero = -196.074 ^{0}C



















8 0
3 years ago
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