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2 years ago

Draw a lewis structure for so2 in which all atoms obey the octet rule. show formal charges. do not consider ringed structures.

Chemistry

1 answer:

Aleonysh [2.5K]2 years ago

6 0

Formal charge = valence electron - bonds - dots.
double bonded oxygen = 6-2-
4 so it is 0
single bonded oxygen = 6-1-6 so it is -1
sulfur = 6-3-2 so it is +1

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Rubbing alcohol contains 615g of isopropanol (C3H7OH) per liter (aqueous solution). Calculate the molality of this solution. Giv

faltersainse [42]

Answer:

Solution of isopropanol is 10.25 molal

Explanation:

615 g of isopropanol (C3H7OH) per liter

We gave the information that 615 g of solute (isopropanol) are contained in 1L of water. We need to find out the mass of solvent, so we use density.

Density of water 1g/mL → Density = Mass of water / 1000 mL of water

Notice we converted the L to mL

Mass of water = 1000 g (which is the same to say 1kg)

Molality are the moles of solute in 1kg of solvent, so let's convert the moles of isopropanol → 615 g . 1mol / 60g = 10.25 moles

Molality (mol/kg) = 10.25 moles / 1kg = 10.25 m

4 0

2 years ago

Determine the percent yield for the reaction between 6.92 g K and 4.28g of oxygen gas if 7.36 g of potassium oxide is produced.

Illusion [34]

Answer:

percentage yield = 88.25%

Explanation:

Firstly, write the chemical reaction and balance the equation.

Potassium react with oxygen to produce potassium oxide.

K + 02 → K2O

Balance the equation

4K + 02 → 2K2O

The limiting reactant is K so the yield of potassium oxide can be calculated using grams for potassium.

atomic mass of K = 39.1g/mol

grams for 4 mole of potassium = 4(39.1) = 156.4 g

grams for 2 moles of K2O = 2( 39.1 × 2 + 16) = 188.4 g

If 156.4 g of K produces 188.4 g of K2O

6.92 g of K will produce ? gram of K2O

cross multiply

grams of K2O = 6.92 × 188.4/156.4

grams of K2O = 1303.72/156.4

grams of K2O = 8.33585677749

grams of K2O = 8.34 g

percentage yield = actual yield/theoretical yield × 100

actual yield = 7.36 g

theoretical yield = 8.34 g

percentage yield = 7.36/8.34 × 100

percentage yield = 736/8.34

percentage yield = 88.2494004796%

percentage yield = 88.25%

5 0

2 years ago

Consider the following hypothetical reaction: 2 P + Q â†’ 2 R + S The following mechanism is proposed for this reaction: P + P Q

aleksandr82 [10.1K]

Answer: the answer is option (D). k[P]²[Q]

Explanation:

first of all, let us consider the reaction from the question;

2P + Q → 2R + S

and the reaction mechanism for the above reaction given thus,

P + P ⇄ T (fast)

Q + T → R + U (slow)

U → R + S (fast)

we would be applying the Rate law to determine the mechanism.

The mechanism above is a three step process where the slowest step seen is the rate determining step. From this, we can see that this slow step involves an intermediate T as reactant and is expressed in terms of a starting substance P.

It is important to understand that laws based on experiment do not allow for intermediate concentration.

The mechanism steps for the reactions in the question are given below when we add them by cancelling the intermediates on the opposite side of the equations then we get the overall reaction equation.

adding this steps gives a final overall reaction reaction.

2P + Q ------------˃ 2R + S

Thus the rate equation is given as

Rate (R) = K[P]²[Q]

cheers, i hope this helps

3 0

2 years ago

Give two areas where the compressible nature of gas is applied

galina1969 [7]

Answer:

1. Gases can be easily liquefied into very small volumes and stored in liquid form Eg in LPGA cylinders and used in homes.

2. Balloons can be easily filled with air.

6 0

1 year ago

What is the molality of sodium chloride in solution that is 13.0% by mass sodium chloride and that has a density of 1.10 g/ml?

8090 [49]

Answer is: molality od sodium chloride is 2,55 mol/kg.
V(solution) = 100 ml.
m(solution) = d(solution) · V(solution).
m(solution) = 1,10 g/ml · 100 ml.
m(solution) = 110 g.
ω(NaCl) = 13,0% = 0,13.
m(NaCl) = ω(NaCl) · m(solution).
m(NaCl) = 0,13 · 110 g.
m(NaCl) = 14,3 g.
n(NaCl) = m(NaCl) ÷ M(NaCl).
n(NaCl) = 14,3 g ÷ 58,5 g/mol.
n(NaCl) = 0,244 mol.
m(H₂O) = 110 g - 14,3 g.
m(H₂O) = 95,7 g = 0,0957 kg.
b(NaCl) = n(NaCl) ÷ m(H₂O).
b(NaCl) = 0,244 mol ÷ 0,0957 kg.
b(NaCl) = 2,55 mol/kg.

3 0

2 years ago