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tatyana61 [14]
2 years ago
11

An unknown element, x, reacts with lithium to form the compound li2x. in other compounds this element also can accommodate up to

12 electrons rather than the usual octet. what element could x be?
Chemistry
2 answers:
Anika [276]2 years ago
8 0

Answer:

X could be S, Se, Te, Po or Lv.

Explanation:

From the formula Li_{2} X, it can be determined the charge of the ion X:

Li is a metal from the group 1 (alkaline metal) in the periodic table,so the only charge possible for it is +1.  As the formula contains two Li, the total positive charge gives +2. This impplies that X must holds a charge of -2.

Inspection of the periodic table shows that atoms that commonly gain two electrons forming -2 charged anions are the ones in group 16.

The second part of the text  states that X can accommodate 12 electrons expanding its octect. This effect can be observed in atoms that have empty d-orbitals available to accomodate those extra electrons. Thus, atoms from period 3 and above show octect expansion.

The elements that are in group 16 in period 3 or above are S, Se, Te, Po and Lv.

aliina [53]2 years ago
7 0
<span>Answer Choices: 
A) Ca 
B) O 
C) Cl 
D) s</span>

<span>the formula is Li2X, so the charge on the X anion must be 2- 

the ion is X2- elements in group 6A form monatomic ions with a 2- charge. In your list that is O or Po 

If the element can accomodate 12 electrons then it can have an expanded octet. Only elements in period 3 and higher can have expanded octets. 

So you are looking for a group 6A element in period 3 or higher.
cA would not intereact with LI BECAUSE ITS A METAL The only element that fits the bill is D) S</span>

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A 0.89% (w/v) sodium chloride solution is referred to as physiological saline solution because it has the same concentration of
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A student pours exactly 26.9 mL of HCl acid of unknown molarity into a beaker. The student then adds 2 drops of the indicator an
Assoli18 [71]
a.
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Hence, HCl reacts with NaOH and gives NaCl salt and H₂O as the products. The reaction is,
            HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l)

To balance the reaction equation, both sides hould have same number of elements.

Left hand side,                                             Right hand side,
             
H atoms = 2                                               H atoms = 2
            Cl atoms = 1                                               Cl atoms = 1
            Na atoms = 1                                               Na atoms = 1 
           O atoms = 1                                                   O atoms = 1

Hence, the reaction equation is already balanced.

b. 
Molarity (M)= moles of solute (mol) / Volume of the solution (L)
 
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Molarity of NaOH = <span>0.13 M
</span>Volume of NaOH added = <span>43.7 mL
Hence, moles of NaOH added = 0.13 M x 43.7 x 10</span>⁻³ L
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Stoichiometric ratio between NaOH and HCl is 1 : 1

Hence, moles of HCl = moles of NaOH
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5.681 x 10⁻³ mol

5.681 x 10⁻³ mol of HCl was in <span>26.9 mL.

Hence, molarity of HCl = </span>5.681 x 10⁻³ mol / 26.9 x 10⁻³ L
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