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VikaD [51]
4 years ago
7

Which is the Lewis structure for H3PO4?

Chemistry
1 answer:
Fed [463]4 years ago
5 0

<u><em>Answer:</em></u>

  • The correct structure of phosphoric acid is A.

<u><em>Explanation</em></u>

  • P should form five covalent bonds. In this strcuture P form three single bond with 3-hydroxyl groups while one single bondformed with oxygen. As oxygen will form two bonds , it carry negative charge, while P should form five bond but here it is forming 4 bonds due to this P has positive charge but overall structure contain neutral charge due to cancellation of positive and negative charges. Beside this, there are 3 H,  Four O and One P according to formula H3PO4.
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Acetone is a common solvent and has a density of 0.7899 g/mL. What volume of acetone, in mL, has a mass of 28.7 g?
Svet_ta [14]
Density = 0.7899 g/mL

mass = 28.7 g

Volume = ?

therefore:

D = m / V

0.7899 =  28.7 / V

V = 28.7 / 0.7899

V  = 36.333 mL




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The process by which an animal becomes accustomed to its situation is referred to as A. Habituation B nurturing C aggression D d
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Many double-displacement reactions are enzyme-catalyzed via the "ping pong" mechanism, so called because the reactants appear to
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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

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}

<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>

7 0
3 years ago
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/~\ The correct answer is:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I hope this helps! /~\

8 0
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