How many molecules are in 210 grams of CO2?

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

4 years ago

Balance each of the following redox reactions occurring in basic solution.MnO−4(aq)+Br−(aq)→MnO2(s)+BrO−3(aq)Express your answer

Ahat [919]

Answer : The balanced chemical equation is,

$2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)$

Explanation :

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.
Now balance the main atoms in the reaction.
Now balance the hydrogen and oxygen atoms on both the sides of the reaction.
If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.
If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion $(OH^-)$ at that side where the less number of hydrogen are present.
Now balance the charge.

The half reactions in the basic solution are :

Reduction : $MnO_4^-(aq)+2H_2O(l)+3e^-\rightarrow MnO_2(s)+4OH^-(aq)$ ......(1)

Oxidation : $Br^-(aq)+6OH^-(aq)\rightarrow BrO_3^-(aq)+3H_2O(l)+6e^-$ .......(2)

Now multiply the equation (1) by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a basic solution will be,

$2MnO_4^-(aq)+Br^-(aq)+H_2O(l)\rightarrow 2MnO_2(s)+BrO_3^-(aq)+2OH^-(aq)$

8 0

3 years ago

Which analogy best represents a supersaturated solution? an empty elevator that has a maximum load of 20 people a single person

Karo-lina-s [1.5K]

25 people in an elevator that has a maximum load of 20 people

8 0

3 years ago

Read 2 more answers

can someone with this? I need someone to do the "THINK" or at least tell me what it's asking for. I Already did the questions in

Jet001 [13]

Answer:

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7 0

3 years ago

An aqueous solution of calcium hydroxide is standardized by titration with a 0.112 M solution of hydrobromic acid. If 15.2 mL of

TiliK225 [7]

Answer:

0.0457 M

Explanation:

The reaction that takes place is:

2HBr + Ca(OH)₂ → CaBr₂ + 2H₂O

First we calculate how many moles of acid reacted, using the HBr solution's concentration and volume:

Molarity = Moles / Volume

Molarity * Volume = Moles

0.112 M * 12.4 mL = 1.389 mmol HBr

Now we convert HBr moles to Ca(OH)₂ moles, using the stoichiometric ratio:

1.389 mmol HBr * $\frac{1mmolCa(OH)_{2}}{2mmolHBr}$ = 0.6944 mmol Ca(OH)₂

Finally we calculate the molarity of the Ca(OH)₂ solution, using the given volume and calculated moles:

0.6944 mmol Ca(OH)₂ / 15.2 mL = 0.0457 M

7 0

3 years ago