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vekshin1
3 years ago
8

What were the limition of doberiner classifacation​

Chemistry
1 answer:
alexgriva [62]3 years ago
5 0

Answer: I don’t kno

Explanation:

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1. Name the compound present in a sample of dry air.
aev [14]

Answer:

not air

Explanation:

I think so

hahaha

7 0
2 years ago
What type of reaction is illustrated?
aev [14]

Answer:

Double replacement is the answer

8 0
3 years ago
If a gas has a volume of 15l and a temperature of 125K, and the temperature is increased to 225K, what is the new volume
aalyn [17]

Answer:

V₂ =  27 L

Explanation:

Given data:

Initial volume = 15 L

Initial temperature = 125 K

Final temperature = 225 K

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 15 L × 225K / 125 k

V₂ = 3375 L.K / 125 K

V₂ =  27 L

6 0
3 years ago
If the value of kc at 25oc is 3.7108, and the equilibrium concentrations for n2 and h2 are 0.000105 m and 0.0000542 m, respectiv
frez [133]

Equation of decomposition of ammonia:

N2+3H2->2NH3

Euilibrium constant:

Kc=(NH3)^2/((N2)((H2)^3))

As concentration of N2=0.000105, H2=0.0000542

so equation will become:

3.7=(NH3)^2/(0.000105)*(0.0000542)^3

NH3=√(3.7*0.000105*(0.0000542)^3)

NH3=7.8×10⁻⁹

So concentration of ammonia will be 7.8×10⁻⁹.

5 0
3 years ago
UNDERSTANDING CONCEPTS-Fill in the Blanks:
igomit [66]

Answer:

17 decrease

18 increase

19 increase

20 decrease

Explanation:

6 0
3 years ago
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