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andrezito [222]
3 years ago
14

What polyatomic ion forms a neutral compound when combined with a group 2A monatomic ion in a 1:1 ratio?

Chemistry
2 answers:
Vinil7 [7]3 years ago
8 0

Answer:

Cabonate

Explanation:

Hi, this are the ions you listed:

<u>Ammonium:</u> NH_4^+

<u>Carbonate:</u> CO_3^{-2}

<u>Nitrate:</u> NO_3^-

<u>Phosphate:</u> PO_4^{-3}

<u>All group 2A compounds (like Ca, Mg) have a valence number of +2</u>. So:

  • They don't bond with ammonium because they are both +
  • With Cabonate they make neutral 1:1 compound: Ca^{+2}CO_3^{-2} \longrightarrow CaCO_3
  • With Nitrate they make positive 1:1 compound: Ca^{+2}NO_3^- \longrightarrow CaNO_3^+
  • With Phosphate they make negative 1:1 compound: Ca^{+2}PO_4^{-3} \longrightarrow CaPO_4^-

lesya692 [45]3 years ago
3 0
It's B. Carbonate hope this helps
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Answer:

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

Explanation:

According to the question , the equation follows :

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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}

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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>

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Answer:

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