Calculate the work done in Cal when 0-425 g of H2O2 decompose against a pressure of 1 atm at 127-C- 2H2O2 l - O2 g -2H2O l R-2Cal-K mole Give your answer in the form of -x-

The expression for the work done is #160; w= PΔ V = Δ n_gRT= 0.425/34 × 2× 2 × 400 Hence, work done w = 5 Cal ...

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Calculate the work done (in Cal) when 0.425 g of $H_{2} O_{2}$ decompose against a pressure of 1 atm at $127^{\circ} C$ ?
$2 H_{2} O_{2} (l) \to O_{2} (g) + 2 H_{2} O (l) (R = 2 C a l / K m o l e)$
(Give your answer in the form of |x|)

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0.425 g

H_{2} O_{2}

127^{o} C

2 H_{2} O_{2} (ℓ) \to O_{2} (g) + 2 H_{2} O (ℓ) (R = 2 c a l / K m o l e)

H_{2} O_{2}

25^{0} C ?

2 H_{2} O_{2} (l) \to O_{2} (g) + 2 H_{2} O (l)

4.5

g

H_{2} O_{2}

1.0

a t m

25^{o} C -

2 H_{2} O_{2} (l) ⟶ O_{2} (g) + 2 H_{2} O (l)

H_{2} O_{2}

2 H_{2} O_{2} (ℓ) \to O 2 (g) + 2 H_{2} O (ℓ)

Calculate the work done (in J) when 4.5 g of $H_{2} O_{2}$ reacts against a pressure of 1.0 atm at $2 H_{2} O_{2} \to O_{2} + 2 H_{2} O$

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