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Lelechka [254]
3 years ago
7

Aluminum boils at 2467°C. Aluminum’s boiling point in Kelvin is 2194. True or False.

Chemistry
2 answers:
Sergeeva-Olga [200]3 years ago
6 0

Answer:

The given statement is false.

Explanation:

Given that aluminum boils at 2,467°C

Boiling point of aluminum in kelvins is 2,194 K

Temperature in degree Celsius can be converted into Kelvins by relation:

(T)^oC=273 +(T) K

So, boiling point of aluminium in Kelvins will be:

2,467 ^oC=273 +(2,467) K=2,740 K

2,740 K ≠ 2,194 K

But the given value in Kelvins is not coming equal to our calculated value, Hence, the given statement is false.

IceJOKER [234]3 years ago
3 0
False. To get Kelvin add 273 to <span>°C to get your temperature into Kelvin. The correct boiling point of Aluminum in Kelvin is 2740 K.</span>
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FromTheMoon [43]

<u>Answer:</u> The concentration of SO_4^{2-} at equilibrium is 0.00608 M

<u>Explanation:</u>

As, sulfuric acid is a strong acid. So, its first dissociation will easily be done as the first dissociation constant is higher than the second dissociation constant.

In the second dissociation, the ions will remain in equilibrium.

We are given:

Concentration of sulfuric acid = 0.025 M

Equation for the first dissociation of sulfuric acid:

       H_2SO_4(aq.)\rightarrow H^+(aq.)+HSO_4^-(aq.)

            0.025          0.025       0.025

Equation for the second dissociation of sulfuric acid:

                    HSO_4^-(aq.)\rightarrow H^+(aq.)+SO_4^{2-}(aq.)

<u>Initial:</u>            0.025            0.025      

<u>At eqllm:</u>      0.025-x          0.025+x        x

The expression of second equilibrium constant equation follows:

Ka_2=\frac{[H^+][SO_4^{2-}]}{[HSO_4^-]}

We know that:

Ka_2\text{ for }H_2SO_4=0.01

Putting values in above equation, we get:

0.01=\frac{(0.025+x)\times x}{(0.025-x)}\\\\x=-0.0411,0.00608

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