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Julli [10]
4 years ago
14

Which statement is correct regarding the reaction below? 3A + 2B yields C + 2D The rate of formation of D is twice the rate of d

isappearance of A. The rate of disappearance of A is one-third the rate of formation of C. The rate of formation of C is twice the rate of appearance of D. The rate of disappearance of B is twice the rate of appearance of C.
Chemistry
2 answers:
Vedmedyk [2.9K]4 years ago
8 0

Answer:

The correct statements are:

The rate of disappearance of B is twice the rate of appearance of C.

Explanation:

Rate of the reaction is a change in the concentration of any one of the reactant or product per unit time.

3A + 2B → C + 2D

Rate of the reaction:

R=-\frac{1}{3}\times \frac{d[A]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}

-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{1}\times \frac{d[C]}{dt}

-\frac{1}{3}\times \frac{d[A]}{dt}=\frac{1}{2}\times \frac{d[D]}{dt}

The rate of disappearance of B is twice the rate of appearance of C.

\frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{2}\times \frac{d[B]}{dt}

2\times \frac{1}{1}\times \frac{d[C]}{dt}=-\frac{1}{1}\times \frac{d[B]}{dt}

Kruka [31]4 years ago
7 0

Answer: D. The rate of disappearance of B is twice the rate of appearance of C.

Explanation:-

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

3A+2B\rightarrow C+2D

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of A = -\frac{1d[A]}{3dt}

Rate in terms of disappearance of B = -\frac{1d[B]}{2dt}

Rate in terms of appearance of C = \frac{1d[C]}{dt}

Rate in terms of appearance of D = \frac{1d[D]}{2dt}

1. The rate of formation of D is 2 by 3 times the rate of disappearance of A.

\frac{1d[D]}{dt}=-\frac{2d[A]}{3dt}

2. The rate of disappearance of A is three times the rate of formation of C.

-\frac{1d[A]}{dt}=\frac{3d[C]}{dt}

3. The rate of formation of C is half the rate of appearance of D.

\frac{1d[C]}{dt}=\frac{1d[D]}{2dt}

4. The rate of disappearance of B is twice the rate of appearance of C.

-\frac{1d[B]}{dt}=\frac{2d[C]}{dt}

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The 18o-labeled methanol (CH3O*H) will appear in the products side at position b.

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The 18O label will appear at position b in the product as indicated in the image.

This methoxy group in the product formed in position b comes from the 18O-labeled methanol (CH3OH).

While the oxygens at positions a and c in the product come from the unlabeled hemiacetal.

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2 years ago
Which of the following equations is correct for coffee-cup calorimeter?
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g 2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electron
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Answer:

In the above reaction, the oxidation state of tin changes from 2+ to 4+.

10 moles of electrons are transferred in the reaction

Explanation:

Redox reaction is:

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

SnO₂²⁻ → SnO₃²⁻

Tin changes the oxidation state from +2 to +4. It has increased it so this is the oxidation from the redox (it released 2 e⁻). We are in basic medium, so we add water in the side of the reaction where we have the highest amount of oxygen. We have 2 O on left side and 3 O on right side so we add 1 water on the right and we complete with OH⁻ in the opposite side to balance the H.  

SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O <u>Oxidation</u>

BrO₃⁻ →  Br₂

First of all, we have unbalance the bromine, so we add 2 on the BrO₃⁻. We have 6 O in left side and there are no O on the right, so we add 6 H₂O on the left. To balance the H, we must complete with 12OH⁻. Bromate reduces to bromine at ground state, so it gained 5e⁻. We have 2 atoms of Br, so finally it gaines 10 e⁻.

6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻ <u>Reduction</u>

In order to balance the main reaction and balance the electrons we multiply  (x5) the oxidation and (x1) the reduciton

(SnO₂²⁻ + 2OH⁻ → SnO₃²⁻ + 2e⁻ + H₂O) . 5

(6H₂O + 10 e⁻ + 2BrO₃⁻ →  Br₂ + 12OH⁻) . 1

5SnO₂²⁻ + 10OH⁻ + 6H₂O + 10 e⁻ + 2BrO₃⁻ → Br₂ + 12OH⁻ + 5SnO₃²⁻ + 10e⁻ + 5H₂O

We can cancel the e⁻ and we substract:

12OH⁻ - 10OH⁻ = 2OH⁻ (on the right side)

6H₂O - 5H₂O = H₂O (on the left side)

2BrO₃⁻ + 5SnO₂²⁻ + H2O ⇄ 5SnO₃²⁻ + Br₂ + 2OH⁻

6 0
4 years ago
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jeyben [28]

Answer: C2HNO3

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Divide by smallest result:

C = 2

H=1

N=1

O = 3

Empirical formula = C2HNO3

8 0
4 years ago
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