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Veseljchak [2.6K]
3 years ago
10

List the following bonds in order of increasing polarity (ionic character). Give the electronegativity values for each atom to j

ustify your answer. Write each bond next to the entry below using the chemical symbol for each element, and use an arrow to show any polarity. A: carbon-sulfur B: chlorine - chlorine C: fluorine – hydrogen D: silicon - oxygen
Chemistry
1 answer:
barxatty [35]3 years ago
4 0

Answer:

A = B < D < C

C - S

Cl - Cl

F ← H

Si → O

Explanation:

The polarity of a bond increases with the increase in the difference in electronegativity. The dipole moment is represented with an arrow pointing the more electronegative atom.

A: carbon-sulfur

C - S

ΔEN = |EN(C) - EN(S)| = |2.5 - 2.5| = 0

B: chlorine - chlorine

Cl - Cl

ΔEN = |EN(Cl) - EN(Cl)| = |3.0 - 3.0| = 0

C: fluorine – hydrogen

F ← H

ΔEN = |EN(F) - EN(H)| = |4.0 - 2.1| = 1.9

D: silicon - oxygen

Si → O

ΔEN = |EN(Si) - EN(O)| = |1.8 - 3.5| = 1.7

The order of increasing polarity is A = B < D < C.

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7 0
2 years ago
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Trava [24]

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2 years ago
How many grams of H,O are needed to produce 13 g of NaOH? 2 Na,O2 + 2 H20-4 NaOH + O2
fgiga [73]

Answer:

This is a typical stoichiometry question.To answer this question you want to get a relationship between

N

a

2

O and NaOH.

So you can get a relationship between the moles of

N

a

2

O

and moles of NaOH by the concept of stoichiometry.

N

a

2

O +

H

2

O ----------------> 2 NaOH.

According to above balanced equation we can have the stoichiometry relationship between

N

a

2

O and NaOH. as 1:2

It means 1 moles of

N

a

2

O is required to react with one mol of

H

2

O to produce 2 moles of NaOH.

in terms of mass 1 mole of

N

a

2

O has mass 62 g on reaction with water produces 2 moles of NaOH or 80 g of NaOH.

62 g of

N

a

2

O produces 80 g of NaOH.

1g of NaOH is produced from 62/80 g of

N

a

2

O

1.6 x

10

2

g of NaOH will require 62 x 1.6 x

10

2

g / 80 of

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124g of

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3 years ago
Calculate the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature (645oc). assume
baherus [9]
Answer : 7.87 X 10^{-6}.

Explanation : To find the fraction of schottky defects in the given lattice of CsCl,

we use the formula, \frac{N_{s} }{N}} = exp ( \frac{-Q_{s}}{2KT} )

on solving with the given values ,Q_{s}= 1.86 eV and T as 645 + 273 K and rest are the constants.

\frac{N_{s} }{N}} = exp ( \frac{-1.86 eV}{2 X (8.62 X 10^{-5}) X (645 +273)} )

we get the answer as 7.87 X 10^{-6}.
6 0
3 years ago
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