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I am Lyosha [343]
2 years ago
8

The zinc in a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (so th

at the hydrochloric acid can get to the zinc). The reaction between the acid and the zinc is: 2H+(aq)+Zn(s)→H2(g)+Zn2+(aq).
When the zinc in a certain penny dissolves, the total volume of gas collected over water at 25 ∘C was 0.961 L at a total pressure of 748 mmHg .
Chemistry
1 answer:
LekaFEV [45]2 years ago
6 0

Answer:

A penny dissolves in hydrochloric acid if the copper coating is filed down in several spot... ... When The Zinc In A Certain Penny Dissolves, The Total Volume Of Gas ... is filed down in several spots (so that the hydrochloric acid can reach the zinc). The reaction between the acid and the zinc 2H+(aq)+Zn(s)→H2(g)+Zn2+(aq) .

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Elements ending in the electron configurations ns^1 are highly reactive metals . What family does these elements belong to ?
myrzilka [38]

Answer:

<h3>A . Alkali metals</h3>

Explanation:

The highlighted elements of the periodic table belong to the alkali metal element family. The alkali metals are recognized as a group and family of elements. These elements are metals. Sodium and potassium are examples of elements in this family.

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6 0
3 years ago
Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to
zhenek [66]

Answer:

<u>D. It will decrease by a factor of 4</u>

Explanation:

According to the question , the equation follows :

A+B\rightarrow C+D

Rate law : This states the rate of reaction is directly proportional to concentration of reactants with each reactant raised to some power which may or may not be equal to the stoichiometeric coefficient.

Rate\ \alpha [A]^{a}[B]^{b}

r=[A]^{a}[B]^{b}.................(1)

STEP": First, find out the power "a" and "b"

a+b = 3 (because it is given that the reaction follow 3rd order-kinetics)

According to question, <u><em>doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,</em></u>

r' = 2r if [A'] = 2[A]

Here [B] is uneffected means [B']=[B]

hence new rate =

r'=[A']^{a}[B']^{b}

Put the value of [A'] , [B'] and r' in the above equation:

2r=[2A]^{a}[B]^{b}...........(2)

Divide equation (1) by (2) we , get

\frac{2r}{r}=\frac{[2A]^{2}[B]^{b}}{[A]^{a}[B]^{b}}

2= 2(\frac{A}{A})^{a}\times (\frac{B}{B})^{b}

Here A and A cancel each other

B and B cancel each other

We get,

2= 2^{a}\times 1^{b}

1^b = 1 ( power of 1 = 1)

2= 2^{a}

This is possible only when a = 1

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1 + b = 3

b =3 -1  = 2

b = 2

Hence the rate law becomes :

r=[A]^{a}[B]^{b}

<u>r=[A]^{1}[B]^{2}.............(3)</u>

Look in the question now, it is asked to calculate the concentration of [B],if  cut in half

Hence

[B']=1/2[B]

Insert the value of [B'] in equation (3)

r'=[A]^{1}[B']^{2}

r'=[A]^{1}(\frac{1}{2}[B])^{2}

r'=\frac{1}{4}[A]^{1}[B]^{2}............(a)

But

r=[A]^{a}[B]^{b}..............(b)

Compare equation (a) and (b) , we get

new rate r' =

<u>r' = 1/4 r</u>

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