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ANEK [815]
2 years ago
9

As we move across the periodic table of elements, certain trends are evident. The placement of individual elements in the period

ic table allows us to predict many characteristics of elements such as state of matter at room temperature, reactivity, and valence electrons just to name a few. Let's compare the elements in Group #1 and Group #17. Identify the differences in the two families based on their locations in the periodic table. Choose ALL that apply
Question 20 options:

Group #1 elements are metals and Group #17 elements are nonmetals.


Group #1 elements form negative ions while Group #17 form positive ions.


Group #1 elements are reactive while Group #17 elements are non-reactive.


When forming compounds, Group #1 elements lose electrons and Group #17 elements gain electrons.
Chemistry
1 answer:
Soloha48 [4]2 years ago
4 0

Answer:

A and E

Explanation:

A and E are the answers! usa test prep told me.    :L

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20 mL of Ba(OH)2 solution with unknown concentration was neutralized by the addition of 43.89 mL of a .1355 M HCl solution. Calc
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Answer:

Concentration of the barium ions  = [Ba^{2+}] = 0.4654 M

Concentration of the chloride ions  = [Cl^{-}]=0.9308 M

Explanation:

Moles (n)=Molarity(M)\times Volume (L)

Moles of hydrogen chloride = n

Volume of hydrogen chloride solution = 43.89 mL = 0.04389 L

Molarity of the hydrogen chloride  = 0.1355 M

n=0.1355 M\times  0.04389 L=0.005947 mol

Ba(OH)_2+2HCl\rightarrow BaCl_2+2H_2O

According to reaction, 2 moles of HCl reacts with 1 mole of barium hydroxide.

Then 0.05947 moles of HCl will react with:

\frac{1}{2}\times 0.05947 mol=0.029735 mol barium hydroxide

Moles of barium hydroxide = 0.029735 mol

Ba(OH)_2(aq)\rightarrow Ba^{2+}(aq)+2OH^-(aq)

1 mole of barium hydroxide gives 1 mole of barium ion in an aqueous solution. Then 0.029735 moles of barium hydroxide will give:

=1\times 0.029735 mol= 0.029735 mol

Volume of solution after neutralization reaction :

= 20.0 mL + 43.89 mL  = 63.89 mL = 0.06389 L

Concentration of the barium ions =[Ba^{2+}]

[Ba^{2+}]=\frac{0.029735 mol}{0.06389 L}=0.4654 M

Ba(Cl)_2(aq)\rightarrow Ba^{2+}(aq)+2Cl^-(aq)

1 mole of barium chloride gives 1 mole of barium ions and 2 moles of chloride ions in an aqueous solution.

Then concentration of chloride ions will be:

[Cl^-]=2\times [Ba^{2+}]=2\times 0.4654 M=0.9308 M

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