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3 years ago

Pls help me in this:(

Chemistry

1 answer:

antiseptic1488 [7]3 years ago

3 0

Answer:

a) 26.98

b) 3

c) 3* 1.01 = 3.03

d) 3 *16 = 48

e) 26.98 + 3.03 + 48 = 78.01

f) 6.023 * 10²³

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Delvig [45]

Answer:

+ 291.9 kJ

Solution:

The equation given is as;

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = ?

First, as we know the heat of formation of H₂O ₍l₎ is,

H₂ ₍g₎ + 1/2 O₂ ₍g₎ → H₂O ₍l₎ ΔH = - 285.9 kJ

Now, reversing the equation will reverse the sign of heat as,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

Also, we know that,

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

Now, adding last two equations,

H₂O ₍l₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 285.9 kJ

H₂O ₍s₎ → H₂O ₍l) ΔH = + 6.0 kJ

-----------------------------------------------------------------------------

H₂O ₍s₎ → H₂ ₍g₎ + 1/2 O₂ ₍g₎ ΔH = + 291.9 kJ

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3 years ago

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Below is a proposed mechanism for the decomposition of H2O2. H2O2 + I– → H2O + IO– slow H2O2 + IO– → H2O + O2 + I– fast Which of

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Answer: Option (d) is the correct answer.

Explanation:

The given equations as as follows.

$H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}$ (slow)

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$2H_{2}O_{2} + I^{-} + IO^{-} \rightarrow 2H_{2}O + O_{2} + IO^{-} + I^{-}$

So, cancelling the spectator ions then the equation will be as follows.

$H_{2}O_{2} \rightarrow 2H_{2}O + O_{2}$

As, it is known that slow step of a reaction is the rate determining step. Therefore, rate law for the slow step will be as follows.

$H_{2}O_{2} + I^{-} \rightarrow H_{2}O + IO^{-}$

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Hence, the reaction is first order with respect to $[I^{-}]$ and it is also first order reaction with respect to $[H_{2}O_{2}]$ .

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Pls help me in this:(​

Pls help me in this:(