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Diano4ka-milaya [45]
1 year ago
15

Does Q for the formation of 1 mol of NH₃ from H₂ and N₂ differ from Q for the formation of NH₃ from H₂ and 1 mol of N₂? Explain

and give the relationship between the two Q’s.
Chemistry
1 answer:
Anton [14]1 year ago
3 0

Yes, the formation of 1 mol of NH3 from H2 and N2 differ from Q for the formation of NH3 from H2 and 1mol of N2

Solution:

Equation for the formation of 1 mol of NH3

      3/2H2(g)+ 1/2N2(g) → NH3(g)

The Q of this reaction is

Q1 = [product] / [reactant]

​Q1 = [NH 3 ] / [H2]^3/2  [N2]^1/2

​

Equation for the formation of NO3 from 1 mol of N2

      3H 2 (g) + N2(g) → 2NH 3(g)

The Q of this reaction

Q 2  = [product] / [reactant]

​Q 2  = [NH 3 ]^2​ /  [H 2 ]^3 [N 2 ]

The relationship between the two Q's is

       (Q1)^2 = Q2​

To learn more about the Q click the given link

brainly.com/question/12693045

#SPJ4

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When writing an electron configuration for a transition metal, what is the last orbital type to be filled – an s, p, d or f orbi
Black_prince [1.1K]
It would be d.

Reason being said...

the electron configuration normally goes like this...

1s2 2s2 2p6 3s2 3p6 4s2....

until you hit the transition metals..remember those have a special rule..

even though you are in the 4 sublevels for the orbitals ... it goes down 1

Making it 3d..(1,2,3,4,5,6,7,8,9,10)

Going on...

at 5s2 then, 4d1, 4d2, 4d3, 4d4, etc..

at 6s2 then, 5d1, 6d2, 6d3, 6d4, etc..

Thus, D orbital is your answer.
4 0
4 years ago
What is the molecular shape of HCN?<br> bent<br> linear<br> angular<br> trigonal pyramidal
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HCN is a linear molecule.
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Balance the following redox equation using the oxidation-number-change method. Show your work, describing each step.
nexus9112 [7]

2HNO3+3H2S ——>2S+4H2O+2NO

5 0
4 years ago
Six moles of an ideal gas are in a cylinder fitted at one end with a movable piston. the initial temperature of the gas is 27.0c
Alla [95]

We have to know final temperature of the gas after it has done 2.40 X 10³ Joule of work.

The final temperature is: 75.11 °C.

The work done at constant pressure, W=nR(T₂-T₁)

n= number of moles of gases=6 (Given), R=Molar gas constant, T₂= Final temperature in Kelvin, T₁= Initial temperature in Kelvin =27°C or 300 K (Given).

W=2.4 × 10³ Joule (Given)

From the expression,

(T₂-T₁)=\frac{W}{nR}

(T₂-T₁)= \frac{2.40 X 10^{3} }{6 X 8.314}

(T₂-T₁)= 48.11

T₂=300+48.11=348.11 K= 75.11 °C

Final temperature is 75.11 °C.


6 0
3 years ago
Which condition will probably not increase
PilotLPTM [1.2K]

Answer:

i think A.

Explanation:

bc B. makes the molecules hight temp = move faster = more collisions = higher rate, C. it's use is to make reactions rates increase, D. somthung abt more surface area and easier collisions

6 0
3 years ago
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