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Zarrin [17]
1 year ago
11

A reaction that occurs in the internal combustion engine isN₂(g) + O₂(g) ⇄ 2NO(g) (a) Determine ΔH° and ΔS° for the reaction at

298 K.
Chemistry
1 answer:
alex41 [277]1 year ago
7 0

ΔH° and ΔS° for the reaction at 298 K  is 24.8 J/mol.K and 180.58 kJ/mol.

A reaction that occurs in the internal combustion engine is

                  N₂(g) + O₂(g) ⇄ 2NO(g)

The standard molar entropy and enthalpy of formations can be calculated using the values of ΔS°and ΔH° as shown below.

NO, N2, and O2 have entropies of 210.65 J/mol. K, 191.5 J/mol. K, and 130.6 J/mol. K, respectively.

Fill in these values in the formula for standard entropy below.

ΔS° = 2S° NO(g) -(S° N2 (g) +S° O2(g))

ΔS°= 2 (210.65 J/mol K) - (191.5 J/mol K + 205.0 J/mol K)

        = 24.8 J/mol K

Consequently, the typical entropy of reaction is 24.8 J/mol.K

NO, N2, and O2 have enthalpies of 90.29 kJ/mol, 0 kJ/mol, and 0 kJ/mol, respectively.

Replace these values in the common enthalpy formula below.

ΔH° = 2ΔH° f(NO(g)) –[ ΔH° f (N2) (g) + ΔH° f (O2)(g)]

ΔH° = 2 (90.29 kJ/mol) - ((0 kJ/mol) + ((0 kJ/mol))

       = 180.58 kJ/mol

Consequently, the usual enthalpy of reaction is 180.58 kJ/mol

Learn more about ΔH° here:

brainly.com/question/14001272

#SPJ4

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What term expresses the ratio of moles of reactants and products according to the coefficients in the balanced chemical equation
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A) mole ratio

Explanation:            

<em>A mole ratio</em>, also known as a mole-to-mole ratio, <em>is ​the ratio between the amounts in moles of one reactant/product to the moles of the other reactant/product. </em>This ratio is determined considering the coefficients in a balanced chemical equation. This ratio is used in chemical problems as a conversion factor between the compounds involved in the reaction.

I hope you find this information useful and interesting! Good luck!

6 0
3 years ago
Question 3 of 10
PIT_PIT [208]

Answer:

the moluculer formula is the answer

Explanation:

4 0
3 years ago
Compared to the nucleus 5626fe, what is the density of the nucleus 112 48cd?
tia_tia [17]

For a comparison of the nucleus 5626fe, the density of the nucleus 112 48cd is mathematically given as the same.

n(Cd) / n(Fe)=1

<h3>What is the density of the nucleus 112 48cd?</h3>

Generally, the equation for the density  is mathematically given as

d=\frac{A}{4/3}\piR^3

Therefore

n(Cd) / n(Fe) = [A (Cd) / (A Fe) ] * [ R (Fe) / R (Cd)]^3

n(Cd) / n(Fe)= (112 / 56 ) * (1/1.26)3

n(Cd) / n(Fe)=1

In conclusion, The ratio of n(Cd) = n(Fe) is 1, hence same

Read more about density

brainly.com/question/14010194

8 0
2 years ago
Can anyone help me on this please?
Bad White [126]
This question is testing to see how well you understand the "half-life" of radioactive elements, and how well you can manipulate and dance around them.  This is not an easy question.

The idea is that the "half-life" is a certain amount of time.  It's the time it takes for 'half' of the atoms in any sample of that particular unstable element to 'decay' ... their nuclei die, fall apart, and turn into nuclei of other elements.

Look over the table.  There are 4,500 atoms of this radioactive substance when the time is 12,000 seconds, and there are 2,250 atoms of it left when the time is ' y ' seconds.  Gosh ... 2,250 is exactly half of 4,500 !  So the length of time from 12,000 seconds until ' y ' is the half life of this substance !  But how can we find the length of the half-life ? ? ?

Maybe we can figure it out from other information in the table !

Here's what I found:

Do you see the time when there were 3,600 atoms of it ? 
That's 20,000 seconds.

... After one half-life, there were 1,800 atoms left.
... After another half-life, there were 900 atoms left.
... After another half-life, there were 450 atoms left. 

==>  450 is in the table !  That's at 95,000 seconds.

So the length of time from 20,000 seconds until 95,000 seconds
is three half-lifes.

The length of time is (95,000 - 20,000) = 75,000 sec

                                     3 half lifes = 75,000 sec

Divide each side by 3 :   1 half life = 25,000 seconds

There it is !  THAT's the number we need.  We can answer the question now.

==> 2,250 atoms is half of 4,500 atoms.

==> ' y ' is one half-life later than 12,000 seconds

==> ' y ' = 12,000 + 25,000

         y   = 37,000 seconds  .

Check: 
Look how nicely 37,000sec fits in between 20,000 and 60,000 in the table.

As I said earlier, this is not the simplest half-life problem I've seen.
You really have to know what you're doing on this one.  You can't
bluff through it.


7 0
4 years ago
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