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konstantin123 [22]
3 years ago
7

Oh no... not again... Prof. Vitarelli spots Sybil running down the hall... yelling something... something about her tea cups...

Prof. Vitarelli and Prof. Snape try to hold her back. But they fail. She yells, I’ve learned another exam question from last year! Zinc metal reacts with hydrochloric acid to produce hydrogen gas and an aqueous solution of zinc(II) chloride. What is the reducing agent in this reaction? Maybe Prof. Umbridge was right in trying to remove her.
Chemistry
1 answer:
Y_Kistochka [10]3 years ago
6 0

Answer:

The reducing agent is Zn.

Explanation:

Let's consider the reaction between zinc and hydrochloric acid.

Zn(s) + 2 HCl(aq) ⇄ ZnCl₂(aq) + H₂(g)

This is a redox reaction, which can be divided in 2 half-reactions: reduction and oxidation.

In the reduction, H⁺ gains electrons and it is considered the oxidizing agent.

2H⁺ + 2 e⁻ ⇒ H₂

In the oxidation, Zn loses electrons and it is considered the reducing agent.

Zn ⇒ Zn²⁺ + 2 e⁻

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C3H8 +502 + 3CO2 + 4H2O
astra-53 [7]

Answer:

Explanation:

first, you calculate the amount of O2 in moles:

98.0 ÷ 32 = 3.0625

second, the ratio if O2/C3H8 is 5 so you need to calculate O2 in moles with that:

3.0625 ÷ 5 = 0.6125

third, the amount of CO2 in moles also can be calculate by the ratio of C3H8/CO2 which is 3

0.6125 × 3 = 1.8375

then multiply CO2 in moles by its molar mass which is 44 g/mol

1.8375 × 44 = 80.85g

5 0
2 years ago
At 25∘C, the decomposition of dinitrogen pentoxide, N2O5(g), into NO2(g) and O2(g) follows first-order kinetics with k=3.4×10−5
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Answer:

4600s

Explanation:

2N_{2}O_{5}(g) - - -> 4NO_{2}(g) +O_{2}

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as:

-\frac{d[B]}{dt}=k[B] - - -  -\frac{d[B]}{[B]}=k*dt

If we have an ideal gas in an isothermal (T=constant) and isocoric (v=constant) process.

PV=nRT we can say that P = n so we can express the reaction order as a function of the Partial pressure of one component.  

-\frac{d[P(B)]}{P(B)}=k*dt  

-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=k*dt

Integrating we get:

\int\limits^p \,-\frac{d[P(N_{2}O_{5})]}{P(N_{2}O_{5})}=\int\limits^ t k*dt

-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])=k(t_{2}-t_{1})

Clearing for t2:

\frac{-(ln[P(N_{2}O_{5})]-ln[P(N_{2}O_{5})_{o})])}{k}+t_{1}=t_{2}

ln[P(N_{2}O_{5})]=ln(650)=6.4769

ln[P(N_{2}O_{5})_{o}]=ln(760)=6.6333

t_{2}=\frac{-(6.4769-6.6333)}{3.4*10^{-5}}+0= 4598.414s

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Answer: D

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