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gtnhenbr [62]
3 years ago
6

Which element in period 3 has three times the electronegativity value compared to that of li in period 2

Chemistry
2 answers:
victus00 [196]3 years ago
7 0

Answer:

Nitrogen

Explanation:

Elements in period two includes lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.

According to periodic trends, the electro negativity values are expected to increase across the period up to fluorine. Hence, as we go right wards, we encounter elements with higher electronegative values.

While lithium has an electronegative value of 1 , the electronegative value of element nitrogen is thrrr times this which is equal to three

qaws [65]3 years ago
4 0

Answer:

Chlorine(3.16)

Explanation:

The electronegativity of an element is the ability of an atom to attract a bonding pair of an atom. This measures the tendency of an atom to attract electron to itself in a chemical bond  . One of the most electronegative element is fluorine.  

Electronegativity in the periodic table increases from left to right across the period while it decreases down the group in the periodic table.

Lithium is in period 2 with electronegativity of 0.98 . The element in period 3 with electronegativity times 3 of lithium is chlorine . Chlorine has an electronegativity of  3.16.

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In logic and philosophy, a property is a characteristic of an object; a red object is said to have the property of redness. The property may be considered a form of object in its own right, able to possess other properties.

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3 years ago
A 3.452 g sample containing an unknown amount of a Ce(IV) salt is dissolved in 250.0-mL of 1 M H2SO4. A 25.00 mL aliquot is anal
SOVA2 [1]

Answer:

1,812 wt%

Explanation:

The reactions for this titration are:

2Ce⁴⁺ + 3I⁻ → 2Ce³⁺ + I₃⁻

I₃⁻ + 2S₂O₃⁻ → 3I⁻ + S₄O₆²⁻

The moles in the end point of S₂O₃⁻ are:

0,01302L×0,03428M Na₂S₂O₃ = 4,463x10⁻⁴ moles of S₂O₃⁻. As 2 moles of S₂O₃⁻ react with 1 mole of I₃⁻, the moles of I₃⁻ are:

4,463x10⁻⁴ moles of S₂O₃⁻×\frac{1molI_{3}^-}{2molS_{2}O_{3}^-} = 2,2315x10⁻⁴ moles of I₃⁻

As 2 moles of Ce⁴⁺ produce 1 mole of I₃⁻, the moles of Ce⁴⁺ are:

2,2315x10⁻⁴ moles of I₃⁻×\frac{2molCe^{4+}}{1molI_{3}^-} = 4,463x10⁻⁴ moles of Ce(IV). These moles are:

4,463x10⁻⁴ moles of Ce(IV)×\frac{140,116g}{1mol} = <em>0,0625 g of Ce(IV)</em>

As the sample has a 3,452g, the weight percent is:

0,0625g of Ce(IV) / 3,452g × 100 = <em>1,812 wt%</em>

I hope it helps!

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