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agasfer [191]
2 years ago
14

0. When measuring tert-butyl alcohol for this experiment, a student first weighs an empty graduated cylinder, then pours 15 mL o

f the alcohol into the graduated cylinder and weighs the cylinder again. He records the amount of alcohol used as the difference in these two masses. What is wrong with this method
Chemistry
1 answer:
lozanna [386]2 years ago
6 0

Answer:

Both have solutions in the  graduated cylinder.

Explanation:

Recording the amount of alcohol used as the difference between two masses is the wrong method used for measuring tert-butyl alcohol for the experiment.  For measuring tert-butyl alcohol for this experiment, the student has to measure the two masses when both the graduated cylinders has solution of tert-butyl alcohol not when one of it is empty (having no tert-butyl alcohol ).

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At what temp will a gas be at if you allow it to expand from an original 456 mL to 65°C to 3.4 L
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What is a dominant trait and a recessive trait
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The ideal gas heat capacity of nitrogen varies with temperature. It is given by:
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Answer:

A)  1059 J/mol

B)  17,920 J/mol

Explanation:

Given that:

Cp = 29.42 - (2.170*10^-3 ) T + (0.0582*10^-5 ) T2 + (1.305*10^-8 ) T3 – (0.823*10^-11) T4

R (constant) = 8.314

We know that:

C_p=C_v+R

We can determine C_v from above if we make C_v the subject of the formula as:

C_v=C_p-R

C_V = 29.42-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4-8.314

C_V = 21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4

A).

The formula for calculating change in internal energy is given as:

dU=C_vdT

If we integrate above data into the equation; it implies that:

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T+(5.82*10^{-7})T2-(1.305*10^{-8})T3-(8.23*10^{-12})T4\,) du

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 1059J/mol

Hence, the internal energy that must be added to nitrogen in order to increase its temperature from 450 to 500 K = 1059 J/mol.

B).

If we repeat part A for an initial temperature of 273 K and final temperature of 1073 K.

then T = 273 K & T2 = 1073 K

∴

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})T/1+(5.82*10^{-7})T2/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1=\int\limits^{500}_{450}(21.106-(2.7*10^{-3})273/1+(5.82*10^{-7})1073/2-(1.305*10^{-8})T3/3-(8.23*10^{-12})T4/4\,)

U2-U1= 17,920 J/mol

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