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siniylev [52]
3 years ago
5

The electron configuration of an element, X, is [He]3s23p1 . The formula of the most probable ionic compound that this element w

ill form with O is The electron configuration of an element, X, is [He]3s23p1 . The formula of the most probable ionic compound that this element will form with O is:_______.
a) X3O2
b) X2O
c) XO
d) X2O3
Chemistry
1 answer:
Yanka [14]3 years ago
3 0

Answer:

Option d) X2O3

Explanation:

The following data were obtained from the question:

The electronic configuration of the element X => [He]3s23p1

Now, let us write the electronic configuration of oxygen. This is given below:

O(8) => 1s2 2s2 2p4

Considering the electronic configuration of the element X, we can the element has 3 electrons in the outermost shell. Oxygen on the other hand has 2 electrons in the outermost shell. During bonding, there will be exchange of ion as shown below:

X³ + O²¯—> X2O3

So, the formula of the most probable ionic compound that this element will form with O is X2O3

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How many grams of Ca(OH)2are required to make 1.5 L of a 0.81 M solution?
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Answer:

Mass = 90.28 g

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Mass of Ca(OH)₂ = ?

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Molarity = number of moles / volume in L

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Number of moles = 0.81 M × 1.5 L

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A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
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Ag_2SO_4

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Formula for the calculation of no. of Mol is as follows:

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mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

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Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

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Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

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