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otez555 [7]
3 years ago
11

Anna stated that ionic compounds have high melting points and low boiling points. She said that this demonstrates that ionic com

pounds have strong bonds because a lot of energy is needed to break the electrical forces that hold the bonds together.
Which statement best describes Anna’s error?

a.Strong electrical forces will make ionic compounds have high boiling points.

b.Strong electrical forces will make ionic compounds have weak bonds.

c.A low amount of energy is needed to break the bonds in ionic compounds.

d.Gravitational forces hold the bonds together in ionic compounds.
Chemistry
2 answers:
Ainat [17]3 years ago
8 0

Answer:

The answer to your question is: letter A

Explanation:

a.Strong electrical forces will make ionic compounds have high boiling points.  This option is correct because ionic bonds are strong which makes substances have high boiling points.

b.Strong electrical forces will make ionic compounds have weak bonds.  This option is false, ionic compounds have strong bonds.

c.A low amount of energy is needed to break the bonds in ionic compounds.  The high amount of energy is needed to break ionic bonds.

d.Gravitational forces hold the bonds together in ionic compounds. Gravitational forces do not affect the ionic compounds.

defon3 years ago
7 0
It’s A cause it’s strong electrical force will make the ionic
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A 0.5438 g of a C.H.O. compound was combusted in air to make 1.039 g of CO2 and 0.6369 g H20. What is the empirical formula? Bal
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Answer:

C₃H₅O₂

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

Explanation:

The reaction can be expressed as:

CₓHₓOₓ + nO₂ → CO₂ + H₂O

Under the assumption that there was a total combustion, all of the carbon in the reactant was combusted into CO₂, so <u>the mass of C contained in the C.H.O. compound is the same mass of C contained in 1.039 g of CO₂</u>:

1.039gCO_{2}*\frac{1molCO_{2}}{44gCO_{2}} *\frac{1molC}{1molCO_{2}} *\frac{12gC}{1molC} =0.2834gC

All of the hydrogens atoms in the compound ended up becoming H₂O, so <u>the mass of H contained in the C.H.O. compound is the same mass of H contained in 0.6369 g of H₂O</u>:

0.6369g*\frac{1molH_{2}O}{18gH_{2}O} *\frac{1molH}{1molH_{2}O} *\frac{1gH}{1molH} =0.0354gH

Because the compound is composed only by C, H and O, <u>the mass of O in the compound can be calculated by substraction</u>:

0.5438 g Compound - 0.2834 g C - 0.0354 g H = 0.2250 g O

In order to determine the empirical formula, we calculate the moles of each component:

  • mol C = 0.2834 g C ÷ 12 g/mol = 0.0236 mol C
  • mol H = 0.0354 g H ÷ 1 g/mol = 0.0354 mol H
  • mol O = 0.2250 g O ÷ 16 g/mol = 0.0141 mol O

Then we divide those values by the lowest one:

0.0236 mol C ÷ 0.0141 = 1.67

0.0354 mol H ÷ 0.0141 = 2.51

0.0141 mol O ÷ 0.0141 = 1

If we multiply those values by 2, we're left with the empirical formula C₃H₅O₂.

  • The reaction is:

4C₃H₅O₂ + 13O₂ → 12CO₂ + 10H₂O

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