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3 years ago

A student wants to see how temperature affects the

Chemistry

2 answers:

Musya8 [376]3 years ago

5 0

Answer:

To monitor Pressure increase

Explanation:

the more pressure, the higher the production of the gas, which can be recorded, more pressure meaning higher production. Then vary the temperature.

Oduvanchick [21]3 years ago

5 0

Answer: Logical, Because it allows the student to catch any invisible gases given off. I hope this helps

Explanation:

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How many joules of energy are absorbed when 36.2 grams of water is evaporated?

AfilCa [17]

Answer:

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Explanation:

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3 years ago

A hot air ballon will rise in air because:

solmaris [256]

Answer: C

Explanation:

Hot air is a lot less dense for a few reasons. Hot air essentially means the particles have more kinetic energy, and move around a lot more. Cold air is dense because the particles move a lot less, have less energy, and are closer together.

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2 years ago

Read 2 more answers

Indicate the FULL NAME and TOTAL NUMBER of each element in the chemical formula provided. *

REY [17]

Answer:

4 lead = Pb

2 nitrogen = N

6 oxygen = O

Explanation:

Know the rules of multiplying wth perentheses.

- Hope that helped! Please let me know if you need further explanation.

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3 years ago

The ideal gas heat capacity of nitrogen varies with temperature. It is given by:

hammer [34]

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

∴

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

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

$U2-U1= 17,920 J/mol$

3 0

3 years ago

You mix 145 grams of water with 200 grams of ethanol to make a solution. What is the solute and what is the solvent?

hodyreva [135]

The greater amount is the solvent and the lesser amount is the solute.

Hence ethanol(200g) which is the greater amount is the solvent here.

And water (145g) which is lesser is the solute here.

7 0

3 years ago