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3 years ago

Draw Lewis structures for each of the following moecules. Show resonance structures, if they exist

1 answer:

6 0

A Lewis structure is a visual representation of the bonds between atoms and it shows the lone pairs of electrons in molecules. This structure is also referred as Lewis dot diagram. Resonance structure are multiple Lewis structure that describe a single molecule.

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The rock in a lead ore deposit contains 89 % PbS by mass. How many kilograms of the rock must be processed to obtain 1.5 kg of P

Zolol [24]

Answer:

Approximately 1.9 kilograms of this rock.

Explanation:

Relative atomic mass data from a modern periodic table:

Pb: 207.2;
S: 32.06.

To answer this question, start by finding the mass of Pb in each kilogram of this rock.

89% of the rock is $\rm PbS$ . There will be 890 grams of $\rm PbS$ in one kilogram of this rock.

Formula mass of $\rm PbS$ :

$M(\mathrm{PbS}) = 207.2 + 32.06 = 239.26\; g\cdot mol^{-1}$ .

How many moles of $\rm PbS$ formula units in that 890 grams of $\rm PbS$ ?

$\displaystyle n = \frac{m}{M} = \rm \frac{890}{239.26} = 3.71980\; mol$ .

There's one mole of $\rm Pb$ in each mole of $\rm PbS$ . There are thus $\rm 3.71980\; mol$ of $\rm Pb$ in one kilogram of this rock.

What will be the mass of that $\rm 3.71980\; mol$ of $\rm Pb$ ?

$m(\mathrm{Pb}) = n(\mathrm{Pb}) \cdot M(\mathrm{Pb}) = \rm 3.71980 \times 207.2 = 770.743\; g = 0.770743\; kg$ .

In other words, the $\rm PbS$ in 1 kilogram of this rock contains $\rm 0.770743\; kg$ of lead $\rm Pb$ .

How many kilograms of the rock will contain enough $\rm PbS$ to provide 1.5 kilogram of $\rm Pb$ ?

$\displaystyle \frac{1.5}{0.770743} \approx \rm 1.9\; kg$ .

5 0

3 years ago

What is the overall equation for the covalent bond in H2O . Is it H2 + O2=H2O OR. H2 + O = H2O

AlexFokin [52]

Answer:

The answer is in the photo

Explanation:

I hope that is useful for you :)

4 0

3 years ago

A solution is made by adding 35.5 mL of concentrated hydrochloric acid ( 37.3 wt% , density 1.19 g/mL1.19 g/mL ) to some water i

erastova [34]

Answer:

1.73 M

Explanation:

We must first obtain the concentration of the concentrated acid from the formula;

Co= 10pd/M

Where

Co= concentration of concentrated acid = (the unknown)

p= percentage concentration of concentrated acid= 37.3%

d= density of concentrated acid = 1.19 g/ml

M= Molar mass of the anhydrous acid

Molar mass of anhydrous HCl= 1 +35.5= 36.5 gmol-1

Substituting values;

Co= 10 × 37.3 × 1.19/36.5

Co= 443.87/36.6

Co= 12.16 M

We can now use the dilution formula

CoVo= CdVd

Where;

Co= concentration of concentrated acid= 12.16 M

Vo= volume of concentrated acid = 35.5 ml

Cd= concentration of dilute acid =(the unknown)

Vd= volume of dilute acid = 250ml

Substituting values and making Cd the subject of the formula;

Cd= CoVo/Vd

Cd= 12.16 × 35.5/250

Cd= 1.73 M

7 0

3 years ago

Julia did an experiment to study the solubility of two substances. She poured 100 mL of water at 20 °C into each of two beakers

Aleks [24]

Answer: The solubility of B is high than the solubility of A.

Explanation:

The solubility is defined as the amount of substance dissolved in a given amount of solvent. More the solute gets dissolved, high will be the solubility and less the solute dissolved, low will be the solubility.

For the given observations:

Mass of undissolved substance of substance A is more than Substance B at every temperature. This implies that less amount of solute gets dissolved in the given amount of solvent.

Hence, substance B has high solubility than substance A.

7 0

3 years ago

Read 2 more answers

Match each atomic particle with the correct charge.

saul85 [17]

Protin 1
Neutron 0
Electron minus 1

7 0

2 years ago

Read 2 more answers