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miv72 [106K]
3 years ago
12

Why do molecular compounds have low melting points and low boiling points relative to ionic substances?​

Chemistry
1 answer:
kap26 [50]3 years ago
4 0

Answer:

The melting and boiling points of molecular compounds are generally quite low compared to those of ionic compounds. This is because the energy required to disrupt the intermolecular forces between molecules is far less than the energy required to break the ionic bonds in a crystalline ionic compound

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How many grams of Ca(OH)2are required to make 1.5 L of a 0.81 M solution?
FrozenT [24]

Answer:

Mass = 90.28 g

Explanation:

Given data:

Mass of Ca(OH)₂ = ?

Volume of solution= 1.5 L

Molarity of solution = 0.81 M

Solution:

First of all we will calculate number of moles.

Molarity = number of moles / volume in L

by putting values,

0.81 M = Number of moles / 1.5 L

Number of moles = 0.81 M × 1.5 L

Number of moles = 1.22 mol

Mass of Ca(OH)₂ in gram:

Mass = number of moles × molar mass

Mass = 1.22 mol × 74.09 g/mol

Mass = 90.28 g

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3 years ago
In a short paragraph, explain one thing you've learned from the course this year.
exis [7]

Answer:

ive learned about the chart , ive learned about the coolest chemicle reactions and how. the can change colors , ive also learned that chenistrey is all around us since the beginning of life ,and how liquid can turn solid

Explanation:

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3 years ago
A solution is a homogeneous mixture of one or more solutes dissolved in a solvent. A specific volume of solvent is only able to
Rashid [163]
I just got this question; the answer is C: supersaturated. Hope this helps.
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3 years ago
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The two naturally occuring isotopes of antimony are 121Sb (57.21%) and 123Sb (42.79%), with isotopic masses of 120.904 and 122.9
emmasim [6.3K]

Answer:

The average atomic weight = 121.7598 amu

Explanation:

The average atomic weight of natural occurring antimony can be calculated as follows :

To calculate the average atomic mass the percentage abundance must be converted to decimal.

121 Sb has a percentage abundance of 57.21%, the decimal format will be

57.21/100 = 0.5721 . The value is the fractional abundance of 121 Sb .

123 Sb has a percentage abundance of 42.79%, the decimal format will be

42.79/100 = 0.4279. The value is the fractional abundance of 123 Sb .

Next step is multiplying the fractional abundance to it masses

121 Sb = 0.5721 × 120.904 = 69.169178400

123 Sb = 0.4279 × 122.904 = 52.590621600

The final step is adding the value to get the average atomic weight.

69.169178400 + 52.590621600 = 121.7598 amu

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3 years ago
2+4 help please i will fail 6th grade
andrey2020 [161]

Answer:

6

Explanation:

lol

4 0
3 years ago
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