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Kisachek [45]
2 years ago
9

Draw the Lewis structure for the sulfur dioxide molecule. Be sure to include all resonance structures that satisfy the octet rul

e.

Chemistry
1 answer:
trapecia [35]2 years ago
7 0

Answer:

See explanation

Explanation:

We have to remember that the formula of <u>sulfur dioxide</u> is SO_2. So, the "S" atom is the central one and the "O" atoms are in the left and right of S. Additionally, "S" and "O" atom has <u>6 valence electrons each</u>. With this in mind, we can start with a structure in which the left oxygen has 2 lone pairs and a double bond, the S atom can have a double bond a single bond and a lone pair and the right oxygen with 3 lone pairs with a single bond <u>(structure A)</u>. Now, if we move the negative charge in the right  (resonance) to the middle and the double bond to the left oxygen we will obtain <u>structure B</u>. Finally, we can move the negative charge in the left to the middle and we will obtain structure C. (See figure 1).

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Calculate the maximum volume (in mL) of 0.143 M HCl that each of the following antacid formulations would be expected to neutral
Nuetrik [128]

Answer:

a. The maximum volume of 0.143 M HCl required is 154.4 mL.

b. The maximum volume of 0.143 M HCl required is 135.7 mL.

Explanation:

a.

Al(OH)_3+3HCl\rightarrow AlCl_3+3H_2O

Mass of aluminum hydroxide = 350 mg =  0.350 g ( 1mg = 0.001 g)

Moles of aluminum hydroxide = \frac{0.350 g}{78 g/mol}=0.004487 mol

According to reaction ,3 moles of HCl neutralize 1 mole of aluminum hydroxide.Then 0.004487 mole of aluminum hydroxide will be neutralize by :

\frac{3}{1}\times 0.004487 mol=0.01346 mol of HCl.

Mg(OH)_2+2HCl\rightarrow MgCL_2+2H_2O

Mass of magnesium hydroxide = 250 mg =  0.250 g ( 1mg = 0.001 g)

Moles of magnesium hydroxide = \frac{0.250 g}{58 g/mol}=0.004310 mol

According to reaction ,2 moles of HCl neutralize 1 mole of magnesium hydroxide.Then 0.004310  mole of magnesium hydroxide will be neutralize by :

\frac{2}{1}\times 0.004310 mol=0.008621 mol of HCl.

Total moles of HCl required to neutralize both :

0.01346 mol + 0.008621 mol = 0.02208 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}{\text{Volume in Liter}}

V=\frac{0.02208 mol}{0.143 M}=0.1544 L

1 L = 1000 mL

0.1544 L = 154.4 mL

The maximum volume of 0.143 M HCl required is 154.4 mL.

b.

CaCO_3+2HCl\rightarrow CaCl_2+H_2O+CO_2

Mass of calcium carbonate = 970mg =  0.970 g ( 1mg = 0.001 g)

Moles of calcium carbonate = \frac{0.970 g}{100 g/mol}=0.00970 mol

According to reaction ,2 moles of HCl neutralize 1 mole of calcium carbonate.Then 0.00970 mole of calcium carbonate will be neutralize by :

\frac{2}{1}\times 0.00970 mol=0.0194 mol of HCl.

Total moles of HCl required to neutralize calcium carbonate : 0.0194 mol

Molarity of the HCL solution = 0.143 M

Volume of the solution = V

Molarity=\frac{\text{Total moles of HCl}}{\text{Volume in Liter}}

V=\frac{0.0194 mol}{0.143 M}=0.1357 L

1 L = 1000 mL

0.1357 L = 135.7 mL

The maximum volume of 0.143 M HCl required is 135.7 mL.

4 0
3 years ago
The elements found in Group 2A, from top to bottom, include beryllium, magnesium, calcium, strontium, and barium. Which of these
pashok25 [27]
Answer is beryllium.

The electronegativity trend is that it increases from left to right in a period and from bottom to top in a group. This is why fluorine has the highest electronegativity out of all the elements. So since beryllium is at the very top of the group, it has the highest electronegativity
6 0
3 years ago
You performed an experiment to find out how far three different sizes of
eduard

Answer:

B how much farther does the arrow fly when you use a stronger bow

Explanation:

6 0
3 years ago
Suppose a student made a different sodium hydroxide solution using 0.401g of solid sodium hydroxide and 200mL of water. The stud
telo118 [61]

Answer:

0.05 M

Explanation:

Mass of benzoic acid= 0.158g

Volume of benzoic acid= 100 mL

Volume of sodium hydroxide = 27.84mL

Molar mass of benzoic acid= 122g/mol

Number of moles of benzoic acid= 0.158g/122g/mol= 1.3 × 10^-3 moles

C= no of moles/volume

C= 1.3 × 10^-3 moles × 1000/100

C= 0.013M

So;

Volume of acid VA = 100mL

Concentration of acid CA= 0.013M

Volume of Base VB = 27.84mL

Concentration of Base CB= ???

Number of moles of acid NA =1

Number of moles of Base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB= CBVBNA

CB= CAVANB/VBNA

CB= 0.013 × 100 × 1/27.84 × 1

CB= 0.05 M

4 0
3 years ago
How many moles of potassium iodide (KI) are in 48.9 g of potassium iodide (KI) ? ____ moles
wel

Answer:

mass of kI = 74.55 Atomic mass

74.55 g ---> 1 mole

48.9 g --> ?

48.9 / 72

0.7 mole.

In one mole of kI --->

5 0
2 years ago
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