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Marrrta [24]
3 years ago
6

How many milliliters of 1.50 m hcl(aq) are required to react with 5.45 g of an ore containing 32.0% zn(s) by mass?

Chemistry
1 answer:
den301095 [7]3 years ago
3 0

Answer is: 33.34 milliliters of HCl.

Balanced chemical reaction: Zn(s) + 2HCl(aq) → H₂(g) + ZnCl₂(aq).

m(ore) = 5.45 g.

ω(Zn) = 32.0% ÷ 100%.

ω(Zn) = 0.32; mass percentage of zinc in an ore.

m(Zn) = ω(Zn) · m(ore).

m(Zn) = 0.32 · 5.45 g.

m(Zn) = 1.635 g.

n(Zn) = m(Zn) ÷ M(Zn).

n(Zn) = 1.635 g ÷ 65.38 g/mol.

n(Zn) = 0.025 mol; amount of the substance.

From chemical reaction: n(Zn) : n(HCl) = 1 : 2.

n(HCl) = 2 · 0.025 mol = 0.05 mol.

V(HCl) = n(HCl) ÷ c(HCl).

V(HCl) = 0.05 mol ÷ 1.50 mol/L.

V(HCl) = 0.033 L · 1000 mL/L.

V(HCl) = 33.34 mL; volume of hydrochloric acid.

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In which of the following equilibrium systems will an increase in the volume cause the equilibrium to shift away from the reacta
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Answer: 2NOBr(g) ⇌ 2NO(g) + Br2(g)

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2NOBr(g) ⇌ 2NO(g) + Br2(g) is the equillibrium system because there are more moles of products,therefore an increase in the volume of the reaction will shift to the right and produce more moles of products. Also both reactants and products exist in the gaseous state and does not have equal number of moles.

5 0
3 years ago
Sulfonation of benzene has the following mechanism: (1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3 [fast] (2) SO3 + C6H6 → H(C6H5+)SO3− [slow]
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Question is incomplete, complete question is as follows :

Complete Question : .Sulfonation of benzene has the following mechanism:

(1) 2 H2SO4 ⇌ H3O+ + HSO4− + SO3

[fast]

(2) SO3 + C6H6 → H(C6H5+)SO3−

[slow]

(3) H(C6H5+)SO3− + HSO4− → C6H5SO3− + H2SO4

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Frist of all, all the common terms are cancelled out and written the overall reaction.

As we know that the rate depednant step is the slowest step of the reaction, rate law is :

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But the problem is that SO3 cannot be written in the overall rate law because it is an intermediate.

Rate law for synthesis of S03 is as follows :

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Hence when we substitute equation 2 in equation one,

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