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3 years ago

What elemental trend exists as we move down a column on the periodic table?

Chemistry

2 answers:

ella [17]3 years ago

5 0

I think it is A, but I am not sure.

Klio2033 [76]3 years ago

3 0

<u>Answer:</u> The correct answer is chemical reactivity of metals increases.

<u>Explanation:</u>

Chemical reactivity is defined as the tendency of an element to loose of gain electrons.

Metals are the elements which looses electrons and hence, their chemical reactivity will be the tendency to loose electrons. Their chemical reactivity increases as we move from top to bottom in a group. This is so because, the electron get added up in the new shell. So, the electron in the outermost orbital gets far away from the nucleus. And hence, this will be lost easily. Thus, the chemical reactivity gets increases.

Non-metals are the elements which gains electrons and hence, their chemical reactivity will be the tendency to loose electrons. Their chemical reactivity decreases as we move from top to bottom in a group. As, the electrons get added up in the new shell. So, the the electron in the outermost orbital gets far away from the nucleus. And hence, the electron will be difficult to gain. Thus, the chemical reactivity of non-metals decreases.

Hence, the correct answer is chemical reactivity of metals increases.

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Consider the insoluble compound silver bromide , AgBr . The silver ion also forms a complex with ammonia . Write a balanced net

Degger [83]

Answer:

- $AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)$

- $K=1.2x10^{-5}$

Explanation:

Hello,

In this case, by considering the dissolution of silver bromide:

$AgBr(s)\rightleftharpoons Ag^+(aq)+Br^-(aq) \ \ \ Ksp=[Ag^+][Br^-]=7.7x10^{-13}$

And the formation of the complex:

$Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)\ \ \ Kf=\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}=1.6x10^7$

We obtain the balanced net ionic equation by adding the aforementioned equations:

$AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)+Ag^+(aq)\\\\AgBr(s)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-(aq)$

Now, the equilibrium constant is obtained by writing the law of mass action for the non-simplified net ionic equation:

$AgBr(s)+Ag^+(aq)+2NH_3(aq)\rightleftharpoons Ag(NH_3)_2^+(aq)+Br^-+Ag^+\\\\K=[Ag^+][Br^-]*\frac{[Ag(NH_3)_2^+]}{[Ag^+][NH_3]^2}$

So we notice that the equilibrium constant contains the solubility constant and formation constant for the initial reactions:

$K=Ksp*Kf=7.7x10^{-13}*1.6x10^{7}\\\\K=1.2x10^{-5}$

Best regards.

4 0

3 years ago

Suppose that 40 mL of water is poured simultaneously into three containers, each having a different sediment. After 10 seconds,

Brut [27]

Answer:

The answer to your question is Sediment C

Explanation:

Permeability is the property of passing through a material without physically or chemically affecting it.

If a substance is very permeable, it will pass through a material rapidly others substances can be permeable but can not pass through the material quickly.

In the data given, we can conclude that the third substance is the least permeable because less quantity of water has passed through it.

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4 years ago