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adell [148]
3 years ago
15

For the reactionN2(g)+O2(g)?2NO(g)classify each of the following actions by whether it causes a leftward shift, a rightward shif

t, or no shift in the direction of the reaction.a) half oxygenb) double oxygenc) double nitrogend) half nitrogene) double nitrogen monoxidef) half nitrogen monoxide
Chemistry
1 answer:
Soloha48 [4]3 years ago
6 0

Answer: Le Chatleir Principle

Explanation:  The answer of this question can be easily explanied through the Le Chatleir Principle which states that In order to maintain the equilibrium of the reaction, the stress applied to that particular direction will move the equilibrium to the opposite direction in which the stress has been applied.

The given reaction is -

      N_{2}(g)+O_{2}(g)\rightleftharpoons 2NO(g)

If the amount of the oxygen is reduced to half and is being doubled, then the reaction will move in froward direction that is a rightward shift. Same situation will arise in option c and d.

If the amount of nitrogen monoxide is reduced to half or being doubled, the reaction will move towards backward direction that is a leftward reaction.

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5 0
2 years ago
Hello I’m having trouble with this question May you help me? Thank you!
Ostrovityanka [42]

Answer:

Explanation:

Here, we want to state the similarity between the model by Rutherford and the model by Thompson

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11 months ago
i like this guy but i don't think he likes me back. he also goes to school somewhere else, but we see eachother every friday. he
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3 0
3 years ago
Read 2 more answers
A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate
inna [77]

Answer:

a) the wet density of the CL sample is 0.0453 lb/in³

b) the water content in the sample is 65.37%

c) the dry density of the CL sample is 0.0274 lb/in³

Explanation:

Given that;

diameter d = 2.83 in

length L = 6 in

weight m = 1.71 lbs

A piece of clay sample had wet-weight of 140.9 grams  and dry-weight of 85.2 grams

a) wet density of the CL sample

wet density can be expressed as  p = M /v

V is volume of sample which is; π/4×d²×L

so p = M / π/4×d²×L

we substitute

p = 1.71 / (π/4 × (2.83)²× 6

p = 1.71 / 37.741

p = 0.0453 lbs/in³

so the wet density of the CL sample is 0.0453 lb/in³

b)

water content of sample is taken as;

w =  (wet_weight - dry_weight) / dry_weight

we substitute

w = (140.9 - 85.2) / 85.2

w = 55.7 / 85.2

w = 0.6537 = 65.37%

therefore the water content in the sample is 65.37%

c)

dry density of the CL sample

to determine the dry density, we say;

Sd = p / ( 1 + w )

we substitute

Sd = 0.0453 / ( 1 + 0.6537)

Sd = 0.0453 /  1.6537

Sd = 0.0274 lb/in³

therefore the dry density of the CL sample is 0.0274 lb/in³

8 0
3 years ago
What information is needed to determine the energy of an electron in a many-electron atom? a) n. b) l. c) ml. d)ms
mel-nik [20]

Answer:

The correct answer is b.

Explanation:

The quantum number n specifies the energetic level of the orbital, the first level being the one with the least energy. As n increases, the probability of finding the electron near the nucleus decreases and the orbital energy increases.

In the case of atoms with more than one electron, the quantum number l also determines the sublevel of energy in which an orbital is found, within a certain energy level. The value of l is designated by the letters s, p, d, and f.

Have a nice day!

8 0
3 years ago
Read 2 more answers
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