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mario62 [17]
3 years ago
10

Which of the following happens when chlorine forms an ion?

Chemistry
2 answers:
Jlenok [28]3 years ago
7 0

What  happens  when  chlorine  form  an ion  is that  it gains an electron and  has  an octet  in its  outer shell  ( answer  A)


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

<u><em> </em></u>Chlorine is  is in atomic  number  17  in periodic table.

The electron configuration  of chlorine  is    1S2 2S2 2P6 3S2 3P5   or[Ne]3S2 3p5  or  2.8.7.

chlorine therefore  has 7 valence electron therefore it  gain  1 electron  to form Cl- ( ion)

Cl- has  8  electron in its outer  shell (  it  obeys  octet  rule  of eight valence in outer shell.

kotykmax [81]3 years ago
7 0
When chlorine forms an ion it gains an electron and has an octet in its outer shell. The correct answer is A. 
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If 35.0 mL of water in a graduated cylinder is displaced by 8.00 mL
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Answer:

<h2>0.52 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}\\

From the question

volume = final volume of water - initial volume of water

volume = 35 - 8 = 27 mL

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density =  \frac{14}{27}  \\  = 0.518518...

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<h3>0.52 g/mL</h3>

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Adding hydrogen atoms to an unsaturated fatty acid will make it __________.
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The answer is (a.) more solid

 

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Calculate the initial rate for the formation of C at 25 ∘C, if [A]=0.50M and [B]=0.075M.Express your answer to two significant f
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The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

<u>Answer:</u> The initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

<u>Explanation:</u>

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation:

A+2B\rightleftharpoons C

Rate law expression for the reaction:

\text{Rate}=k[A]^a[B]^b

where,

a = order with respect to A

b = order with respect to B

  • Expression for rate law for first trial:

5.4\times 10^{-3}=k(0.30)^a(0.050)^b ....(1)

  • Expression for rate law for second trial:

1.1\times 10^{-2}=k(0.30)^a(0.100)^b ....(2)

  • Expression for rate law for third trial:

2.2\times 10^{-2}=k(0.50)^a(0.050)^b ....(3)

Dividing 2 by 1, we get:

\frac{1.1\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.30)^a(1.00)^b}{(0.30)^a(0.050)^b}\\\\2=2^b\\b=1

Dividing 3 by 1, we get:

\frac{2.2\times 10^{-2}}{5.4\times 10^{-3}}=\frac{(0.50)^a(0.050)^b}{(0.30)^a(0.050)^b}\\\\4.07=2^a\\a=2

Thus, the rate law becomes:

\text{Rate}=k[A]^2[B]^1       ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

5.4\times 10^{-3}=k[0.30]^2[0.050]^1\\\\k=1.2M^{-2}s^{-1}

Calculating the initial rate of formation of C by using equation 4, we get:

k=1.2M^{-2}s^{-1}

[A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get:

\text{Rate}=1.2\times (0.50)^2\times (0.075)^1\\\\\text{Rate}=2.25\times 10^{-2}Ms^{-1}

Hence, the initial rate for the formation of C at 25°C is 2.25\times 10^{-2}Ms^{-1}

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