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LUCKY_DIMON [66]
2 years ago
13

Which of these phrases accurately describe disproportionation? Check all that apply.

Chemistry
2 answers:
Olin [163]2 years ago
7 0

Answer:

A single compound is simultaneously oxidized and reduced.

Explanation:

In chemistry, disproportionation is a simultaneous oxidation and reduction of a single chemical specie.

What this means is that; in a disproportionation reaction, only one compound is both oxidized and reduced. This implies that two products are formed during disproportionation. One is the oxidized product while the other is the reduced product.

Consider the disproportionation of CuCl shown below;

2CuCl -----> CuCl2 + Cu

Here, CuCl2 is the oxidized product while Cu is the reduced product.

stich3 [128]2 years ago
7 0

Answer:

A & c & e

Explanation:

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Which describes the state at which products form at the same rate as reactants?
swat32

Chemical equilibrium<span> is the state in which both reactants and products are present in concentrations which have no further tendency to change with time.
</span><span>Or, we can say that in chemical equilibrium the  ratio between the concentration of the reactants and the products is constant.</span><span>
Chemical equilibrium is a result state when </span><span>the forward reaction proceeds at the same rate as the reverse reaction.
</span><span>Different reactions have different equilibrium.</span>

8 0
3 years ago
Read 2 more answers
Need help ASAP please show your work
VikaD [51]

Answer:

q = 14049 J

Explanation:

q = m*c*(t2-t1)

q = 350 * 0.892 * (70-25) =

312.2 * 45 = 14049 J

I might be getting a little confused but I could be right.

Hope this helps!

4 0
3 years ago
Why is it that 85.48 rounded to two significant figures is 85 and not 86?
Gelneren [198K]

Answer:

See below.

Explanation:

That is because  of the .48.

85.48 is closer to 85 than 86.

8 0
3 years ago
PLZZZZZ HELP MEEEEEE
Masja [62]

Answer:

0.0745 mole of hydrogen gas

Explanation:

Given parameters:

Number of H₂SO₄ = 0.0745 moles

Number of moles of Li = 1.5107 moles

Unknown:

Number of moles of H₂ produced = ?

Solution:

To solve this problem, we have to work from the known specie to the unknown one.

The known specie in this expression is the sulfuric acid,  H₂SO₄. We can compare its number of moles with that of the unknown using a balanced chemical equation.

   Balanced chemical equation:

                    2Li   +     H₂SO₄   →   Li₂SO₄   +   H₂

 From the balanced equation;

     

Before proceeding, we need to obtain the limiting reagent. This is the reagent whose given proportion is in short supply. It determines the extent of the reaction.

           2 mole of Li reacted with 1 mole of  H₂SO₄

          1.5107 mole of lithium will react with \frac{1.5107}{2}  = 0.7554mole of H₂SO₄

But we were given 0.0745 moles,

This suggests that the limiting reagent is the sulfuric acid because it is in short supply;

   

   since 1 mole of sulfuric acid produced 1 mole of hydrogen gas;

    0.0745 mole of sulfuric acid will produce 0.0745 mole of hydrogen gas

4 0
3 years ago
Calculate the pH and fraction of dissociation ( α ) for each of the acetic acid ( CH 3 COOH , p K a = 4.756 ) solutions. A 0.002
marysya [2.9K]

Answer:

The degree of dissociation of acetic acid is 0.08448.

The pH of the solution is 3.72.

Explanation:

The pK_a=4.756

The value of the dissociation constant = K_a

pK_a=-\log[K_a]

K_a=10^{-4.756}=1.754\times 10^{-5}

Initial concentration of the acetic acid = [HAc] =c = 0.00225

Degree of dissociation = α

HAc\rightleftharpoons H^++Ac^-

Initially

c

At equilibrium ;

(c-cα)                                cα        cα

The expression of dissociation constant is given as:

K_a=\frac{[H^+][Ac^-]}{[HAc]}

1.754\times 10^{-5}=\frac{c\times \alpha \times c\times \alpha}{(c-c\alpha)}

1.754\times 10^{-5}=\frac{c\alpha ^2}{(1-\alpha)}

1.754\times 10^{-5}=\frac{0.00225 \alpha ^2}{(1-\alpha)}

Solving for α:

α = 0.08448

The degree of dissociation of acetic acid is 0.08448.

[H^+]=c\alpha = 0.00225M\times 0.08448=0.0001901 M

The pH of the solution ;

pH=-\log[H^+]

=-\log[0.0001901 M]=3.72

3 0
3 years ago
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