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makkiz [27]
3 years ago
6

For most atoms, a stable configuration of electrons is attained when the atom __________. hints hint 1. (click to open) for most

atoms, a stable configuration of electrons is attained when the atom __________. achieves a zero net charge has a completely filled outermost shell has moved all its electrons to its outermost shell has as many protons as neutrons
Chemistry
1 answer:
sveticcg [70]3 years ago
5 0

Correct answer: has a completely filled outermost shell

Atoms of the element with complete outermost shells are stable. So, in order to attain stability the atom either loses electrons or gains electrons to completely fill the outermost shell. The stable electronic configuration for the s and p-block elements is exhibited by the noble gases or the group 8 elements. All the unstable atoms try to attain the electronic configuration of the nearest noble gas with completely filled outermost shell.

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The substance has a higher density than water


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How many moles of N2 in 57.1 g of N2?
SpyIntel [72]

We are given –

  • Mass of \bf N_2 is 57.1 g and we are asked to find number of moles present in 57.1 g of \bf N_2

\qquad\pink{\bf\longrightarrow  { Molar \:mass \:of \: N_2:-}  }

\qquad\bf  \twoheadrightarrow 14\times 2

\qquad\bf \twoheadrightarrow   28

\qquad____________________

Now,Let's calculate the number of moles present in 57.1 g of \bf N_2

\qquad\purple{\bf\longrightarrow  { No \:of \:moles = \dfrac{Given \:mass}{Molar\: mass}}}

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7 0
2 years ago
A detailed explanation, one paragraph of the colligative property being discussed and why that property changes the way that it
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You can do it on the icing of roads, reverse osmosis for desalination of water, dissolved CO2 in soda cans, osmotic pressure involving blood vessels and IV solutions, etc.</span>
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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>

7 0
2 years ago
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MrRa [10]
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