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Verdich [7]
4 years ago
9

Hbr is mixed with water. which is the best description of what you would expect to find in the solution? 1. h2br+, oh1−, and wat

er molecules 2. h3o +, br1−, and water molecules 3. hbr molecules and water molecules 4. hbr molecules, h3o +, br1−, and water molecules 5. br1−, oh1−, and water molecules 6. br+, oh1−, and water molecules
Chemistry
1 answer:
poizon [28]4 years ago
3 0
The correct answer is 2, H3O + Br-1 and water.
Hydrogen bromide is a strong acid, which dissolves totally in water. In solution HBr will be reduced to hydronium ion {H3O} and bromide ion {Br-}.
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Of the molecules below, the bond in __________ is the most polar. hbr hi hcl hf
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For the reaction A + B + C → D + E, the initial reaction rate was measured for various initial concentrations of reactants. The
babymother [125]

Answer : The order of reaction with respect to A is, second order reaction.

The order of reaction with respect to B is, zero order reaction.

The order of reaction with respect to C is, first order reaction.

Explanation :

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+B+C\rightarrow D+E

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b[C]^c

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

6.0\times 10^{-4}=k(0.20)^a(0.20)^b(0.20)^c ....(1)

Expression for rate law for second observation:

1.8\times 10^{-3}=k(0.20)^a(0.20)^b(0.60)^c ....(2)

Expression for rate law for third observation:

2.4\times 10^{-3}=k(0.40)^a(0.20)^b(0.20)^c ....(3)

Expression for rate law for fourth observation:

2.4\times 10^{-3}=k(0.40)^a(0.40)^b(0.20)^c ....(4)

Dividing 1 from 2, we get:

\frac{1.8\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.20)^a(0.20)^b(0.60)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\3=3^c\\c=1

Dividing 1 from 3, we get:

\frac{2.4\times 10^{-3}}{6.0\times 10^{-4}}=\frac{k(0.40)^a(0.20)^b(0.20)^c}{k(0.20)^a(0.20)^b(0.20)^c}\\\\4=2^a\\a=2

Dividing 3 from 4, we get:

\frac{2.4\times 10^{-3}}{2.4\times 10^{-3}}=\frac{k(0.40)^a(0.40)^b(0.20)^c}{k(0.40)^a(0.20)^b(0.20)^c}\\\\1=2^b\\b=0

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^0[C]^1

Thus,

The order of reaction with respect to A is, second order reaction.

The order of reaction with respect to B is, zero order reaction.

The order of reaction with respect to C is, first order reaction.

5 0
3 years ago
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