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3 years ago

Similarities and differences between covalent bonds and molecular interactions

1 answer:

3 0

Covalent bonds are formed between atoms which have
- unsatisfied valency 
- no inert gas electronic configuration 
- These are directional bonds 
- formed by sharing of electrons 

Intermolecular forces 
- much weaker than covalent bond 
- These are not directional (except Hydrogen bonds) 
- These are more electrostatic in nature 
- exist between stable molecules 
- can be Hydrogen bonding, dipole-dipole and induced dipole-induced dipole

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A thermometer reads an outside air temperature of 35 °C. What is the temperature in degrees Fahrenheit?

Rasek [7]

To convert take 1.8 * degrees C + 32
1.8 * 35 + 32
63 + 32 = 95 so C

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3 years ago

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What is the molecular formula of a compound with an empirical formula of CFBrO and a molar mass of 254.7 grams per mole?

miss Akunina [59]

It’s C2F2Br2O2. Can you mark me Brain

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3 years ago

When heat is applied to 80 grams of CaCO3, it yields 39 grams of CO2. Determine

galben [10]

Answer: The percentage yield of $CO_2$ is 90.26%.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ ......(1)

Given mass of $CaCO_3$ = 80 g

Molar mass of $CaCO_3$ = 100 g/mol

Plugging values in equation 1:

$\text{Moles of }CaCO_3=\frac{80g}{100g/mol}=0.8 mol$

For the given chemical equation:

$CaCO_3\rightarow CaO+CO_2$

By the stoichiometry of the reaction:

If 1 mole of $CaCO_3$ produces 1 mole of $CO_2$

So, 0.8 moles of $CaCO_3$ will produce = $\frac{1}{1}\times 0.8=0.8mol$ of $CO_2$

Molar mass of $CO_2$ = 44 g/mol

Plugging values in equation 1:

$\text{Mass of }CO_2=(0.8mol\times 44g/mol)=35.2g$

The percent yield of a reaction is calculated by using an equation:

$\% \text{yield}=\frac{\text{Actual value}}{\text{Theoretical value}}\times 100$ ......(2)

Given values:

Actual value of $CO_2$ = 35.2 g

Theoretical value of $H_2CO_3$ = 39 g

Plugging values in equation 2:

$\% \text{yield of }CO_2=\frac{35.2g}{39g}\times 100\\\\\% \text{yield of }CO_2=90.26\%$

Hence, the percentage yield of $CO_2$ is 90.26%.

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3 years ago

Which is an example of poor safety practices when working outdoors

kati45 [8]

Answer:

Explanation:

u can't touch a chemical with bare skin

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3 years ago

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If the crucible originally weighs 3.715 g and 2. 000 g of hydrate are added to it , what is the weight of the water that is lost

dimulka [17.4K]

For this, you need to know 1) the mass of the hydrate and 2) the mass of the anhydrous salt. Once you have both of these, you will subtract 1) from 2) to find the mass of the water lost.

From the problem, you know that 1) = 2.000 g.

Now you need to find 2). You know that your crucible+anhydrous salt is 5.022 g. To find just the anhydrous salt, subtract the mass of the crucible (3.715 g).

1) = 5.022 g - 3.715 g = 1.307 g

Now you can complete our original task.

Mass H2O = 2) - 1) = 2.000 g - 1.307 g = 0.693 g.

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4 years ago