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3 years ago

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20. Which of the following elements would you expect to lose one electron? Select all that apply.

1 answer:

Jobisdone [24]3 years ago

5 0

Answer:

Sodium and Potassium only

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4 years ago

What is the atomic number of the as yet undiscovered element in which the 8s and 8p electron energy levels?

Andreyy89

The atomic number of the undiscovered element is 168

Element 118 will have just filled its 7p orbitals. therefore the predicted element to fill completely up to its 8 p orbital would have to filled a whole set of s, p, d, f and g orbitals

That's another 2 + 6 + 10 +14 + 18 = 50 electrons

To determine the total number of quantum numbers we have to find

Nml × Nms

we have Nml × Nms = ( 2 + 1 ) × 2

8s + 8P + 7d + 6f + 5g = 2 + 6 + 10 + 14 + 18 = 50

The element right below should be

Z = 118 + 50

= 168

Hence the atomic number of the undiscovered element is 168

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1 year ago

Identify the solute with the highest van't Hoff factor. And how do you determine which one is highest?

german

Answer : The correct option is, (A) $AlCl_3$

Explanation :

Van't Hoff factor : It is defined as the ratio of the experimental value of the colligative property to the calculated value of the colligative property.

We can determine the van't Hoff factor by the association and dissociation of the compound.

(A) $AlCl_3$

It is an electrolyte that dissociates to give aluminum ion and chloride ion.

The dissociation of $AlCl_3$ will be,

$AlCl_3\rightarrow Al^{3+}+3Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Al^{3+}+3Cl^{-}$ = 1 + 3 = 4

(B) $KI$

It is an electrolyte that dissociates to give potassium ion and iodide ion.

The dissociation of $KI$ will be,

$KI\rightarrow K^{+}+I^{-}$

So, Van't Hoff factor = Number of solute particles = $K^{+}+I^{-}$ = 1 + 1 = 2

(C) $CaCl_2$

It is an electrolyte that dissociates to give calcium ion and chloride ion.

The dissociation of $CaCl_2$ will be,

$CaCl_2\rightarrow Ca^{2+}+2Cl^{-}$

So, Van't Hoff factor = Number of solute particles = $Ca^{2+}+2Cl^{-}$ = 1 + 2 = 3

(D) $MgSO_4$

It is an electrolyte that dissociates to give magnesium ion and sulfate ion.

The dissociation of $MgSO_4$ will be,

$MgSO_4\rightarrow Mg^{2+}+SO_4^{2-}$

So, Van't Hoff factor = Number of solute particles = $Mg^{2+}+SO_4^{2-}$ = 1 + 1 = 2

(E) Non-electrolyte

The dissociation non-electrolyte is not possible. So, the Van't Hoff factor will always be, 1.

Hence, the highest van't Hoff factor of solute is, $AlCl_3$

7 0

3 years ago