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3 years ago

What is the oxidation state of S in HSO4^2-? (A.P.E.X)

1 answer:

7 0

I think there is a typo because I've never seen HSO4 2- before in my life. It should be HSO4-. For that, H is 1+ and each Oxygen is 2-0 totaling 8-. So the oxidation state of sulfur +1 - 8 = 7

So the oxidation state of sulfur is +6

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Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to

zhenek [66]

Answer:

D. It will decrease by a factor of 4

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, doubling the concentration of the first reactant causes the rate to increase by a factor of 2 means,

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

We know that : a + b = 3

1 + b = 3

b =3 -1 = 2

b = 2

Hence the rate law becomes :

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

$r=[A]^{1}[B]^{2}$ .............(3)

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' =

r' = 1/4 r

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3 years ago

What is the chemical name for K2Se

Oduvanchick [21]

Answer:

Potassium selenide

Explanation:

Potassium selenide (K2Se)

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How many moles of carbon and hydrogen are needed to produce 10 mol of ethylene

alekssr [168]

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Bess [88]

Alkaline Earth Metals are the elements located in the second period from the left of the periodic table. These elements lose two electrons to form the stable octet when forming an ionic bond, resulting in a net charge of +2. Because they’re trying to get rid of those electrons to get to the stable octet, it’s easy to remove them - this means that the ionization energy of these elements is relatively low. Finally, since they’re looking to get rid of electrons, they certainly aren’t trying to gain any, meaning that their electronegativity is relatively low.

The correct answers are A and D.

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Aleonysh [2.5K]

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