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monitta
3 years ago
7

The branch of chemistry that looks at the release of electrical energy is called

Chemistry
1 answer:
sergij07 [2.7K]3 years ago
3 0
B is the correct answer tbh
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Determine the change in volume that takes place when a 2.00 L sample of N2 gas is heated from 250 C to 500 C.
Effectus [21]

Answer:

V₂ = 2.96 L

Explanation:

Given data:

Initial volume = 2.00 L

Initial temperature =  250°C

Final volume = ?

Final temperature = 500°C

Solution:

First of all we will convert the temperature into kelvin.

250+273 = 523 k

500+273= 773 k

According to Charles's law,

V∝ T

V = KT

V₁/T₁ = V₂/T₂

V₂ = T₂V₁/T₁

V₂ = 2 L × 773 K / 523 k

V₂ =  1546 L.K / 523 k

V₂ = 2.96 L

4 0
3 years ago
Answer the ones please get right
Natali [406]

Answer:

Hope this helps!

Explanation:

I would help you, but I won't. Because you take others people points without helping them and but random words. So, since you are doing that I won't be helping you <3. Thanks for the points idiot.

6 0
3 years ago
Which of the following is an electrolyte?<br> CH4OH<br> O Ca(NO3)2<br> O C12H22011<br> O CH3CH2OH
Slav-nsk [51]

Answer:

O CH3CH2OH

Explanation:

5 0
3 years ago
Why is it possible to compress a gas
dexar [7]

Answer:

Gas can be compress easily than liquid or solid because, molecules are more spread in gas than solin or liquid.

Brainliest if this helps you. please

8 0
2 years ago
Calculate the concentration (M) of sodium ions in a solution made by diluting 50.0 mL of a 0.874 M solution of sodium sulfide to
statuscvo [17]

Answer:

Since one mole Na₂S ionised to give 2moles Na⁺ ion, hence concentration of sodium ion in the solution is (2 × 0.1748)M = 0.3496 M

≈0.350 M

Explanation:

From the question it is clear that,  

Initial volume of sodium sulphide solution is (v₁) = 50mL

Initial concentration of sodium sulphide solution is (s₁) =0.874 M

Final volume of sodium sulphide solution is (v₂) = 250mL

Let, the final concentration of sodium sulphide solution is s₂, then according to acidimetry-alkalimetry,

v₁ × s₁ = v₂ × s₂

Or, s₂ = v₁ × s₁/v₂

= 50 × 0.874 / 250

= 0.1748 M

Therefore, concentration of 250mL sodium sulphide solution is 0.1748 M

Since one mole Na₂S ionised to give 2moles Na⁺ ion, hence concentration of sodium ion in the solution is (2 × 0.1748)M = 0.3496 M

≈0.350 M

5 0
3 years ago
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