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2 years ago

HI(aq)+NaOH(aq)→ what the final balanced chemical equation with the phases included

Chemistry

1 answer:

julia-pushkina [17]2 years ago

7 0

Answer:

$HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)$

Explanation:

Hello there!

In this case, for this neutralization reaction, it is possible to realize that one the neutralization products is water (pH=7) and the other one is the salt coming up from the cation of the NaOH and the anion of the HI:

$HI(aq)+NaOH(aq)\rightarrow NaI+H_2O$

Moreover, since the solubility of NaI is large in water, we infer it remains aqueous whereas the water is maintained as liquid:

$HI(aq)+NaOH(aq)\rightarrow NaI(aq)+H_2O(l)$

Which is also balanced as the number of atoms of all the elements is the same at both sides.

Best regards!

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Answer:

$J_{m} = 20$

$J_{m} = 22.5$

Explanation:

From the question we are told that

The first temperatures is $T_1 = 25^oC = 25 +273 =298 \ K$

The second temperature is $T_2 = 100^oC = 100 +273 = 373 \ K$

Generally the equation for the most highly populated rotational energy level is mathematically represented as

$J_{m} = [ \frac{RT}{2B}] ^{\frac{1}{2} } - \frac{1}{2}$

Here R is the gas constant with value $R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}$

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Generally the energy require per mole to move 1 cm is 12 J /mole

So $0.244 \ cm^{-1}$ will require x J/mole

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So at the second temperature

$J_{m} = [ \frac{8.314 * 373 }{2* 2.928 }] ^{\frac{1}{2} } - 0.5$

=> $J_{m} = 22.5$

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<h3>✽ - - - - - - - - - - - - - - - ~<u>Hello There</u>!~ - - - - - - - - - - - - - - - ✽</h3>

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