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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)

1 answer:

Simora [160]3 years ago

3 0

Answer:

There is an extra O2 molecule left over

Explanation:

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3 years ago

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 = $-196.074 ^{0}C$

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3 years ago