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2 years ago

12.00 moles of naclo3 will produce how many grams of O2? (Moles to Grams)

1 answer:

4 0

2NaClO₃ → 2NaCl + 3O₂

mole ratio of NaClO₃ to O₂ is 2 : 3

∴ if moles of NaClO₃ = 12 mol

then moles of O₂ = $\frac{12 mol * 3}{2}$

= 18 mol

Mass of O₂ = mol of O₂ × molar mass of O₂

= 18 mol × 16 g/mol

= 288 g

So I wasn't sure which equation to use since you did not specify so I just used the decomposition reaction. If you should have used another reaction then just follow the same steps and you'll get your answer.

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What fraction of the earth did the former continental glaciers cover?

bija089 [108]

i think one third part of the earth

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3 years ago

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Which of these is a benefit of nuclear fusion?

dybincka [34]

Answer:

B. There is a lower risk of runaway chain reactions.

Explanation:

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2 years ago

The teacher in item 10 also needs to order plastic tubing. If each of the 60 students need 750 mm of tubing, what length of tubi

Alex777 [14]

45 m. If each student needs 750 mm of tubing, the teacher should order 45 m of tubing.

a) Find the length in millimetres

Length = 60 students x (750 mm tubing/1 student) = 45 000 mm tubing

b) Convert millimetres to metres

Length = 45 000 mm tubing x (1 m tubing/1000 mm tubing) = 45 m tubing

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2 years ago

An analytical chemist is titrating of a solution of propionic acid with a solution of 224.9 ml of a 0.6100M solution of propioni

Svetllana [295]

Answer: The pH of acid solution is 4.58

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}$ .....(1)

For KOH:

Molarity of KOH solution = 1.1000 M

Volume of solution = 41.04 mL

Putting values in equation 1, we get:

$1.1000M=\frac{\text{Moles of KOH}\times 1000}{41.04}\\\\\text{Moles of KOH}=\frac{1.1000\times 41.04}{1000}=0.04514mol$

For propanoic acid:

Molarity of propanoic acid solution = 0.6100 M

Volume of solution = 224.9 mL

Putting values in equation 1, we get:

$0.6100M=\frac{\text{Moles of propanoic acid}\times 1000}{224.9}\\\\\text{Moles of propanoic acid}=\frac{0.6100\times 224.9}{1000}=0.1372mol$

The chemical reaction for propanoic acid and KOH follows the equation:

$C_2H_5COOH+KOH\rightarrow C_2H_5COOK+H_2O$

Initial: 0.1372 0.04514

Final: 0.09206 - 0.04514

Total volume of solution = [224.9 + 41.04] mL = 265.94 mL = 0.26594 L (Conversion factor: 1 L = 1000 mL)

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{salt}]}{[acid]})$

$pH=pK_a+\log(\frac{[C_2H_5COOK]}{[C_2H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of propanoic acid = 4.89

$[C_2H_5COOK]=\frac{0.04514}{0.26594}$

$[C_2H_5COOH]=\frac{0.09206}{0.26594}$

pH = ?

Putting values in above equation, we get:

$pH=4.89+\log(\frac{(0.04514/0.26594)}{(0.09206/0.26594)})\\\\pH=4.58$

Hence, the pH of acid solution is 4.58

7 0

3 years ago

You have a solution of 600 mg of caffeine dissolved in 100 mL of water. The partition coefficient for aqueous caffeine extracted

klio [65]

Answer:

159 mg caffeine is being extracted in 60 mL dichloromethane

Explanation:

Given that:

mass of caffeine in 100 mL of water = 600 mg

Volume of the water = 100 mL

Partition co-efficient (K) = 4.6

mass of caffeine extracted = ??? (unknown)

The portion of the DCM = 60 mL

Partial co-efficient (K) = $\frac{C_1}{C_2}$

where; $C_1=$ solubility of compound in the organic solvent and $C_2$ = solubility in aqueous water.

So; we can represent our data as:

$K=(\frac{A_{(g)}}{60mL} )$ ÷ $(\frac{B_{(mg)}}{100mL} )$

Since one part of the portion is A and the other part is B

A+B = 60 mL

A+B = 0.60

A= 0.60 - B

4.6= $(\frac{0.6-B(mg)}{60mL} )$ ÷ $(\frac{B_{(mg)}}{100mL})$

4.6 = $\frac{(\frac{0.6-B(mg)}{60mL} )}{(\frac{B_{(mg)}}{100mL})}$

4.6 × $(\frac{B_{(mg)}}{100mL})$ = $(\frac{0.6-B(mg)}{60mL} )$

4.6 B $*\frac{60}{100}$ = 0.6 - B

2.76 B = 0.6 - B

2.76 + B = 0.6

3.76 B = 0.6

B = $\frac{0.6}{3.76}$

B = 0.159 g

B = 159 mg

∴ 159 mg caffeine is being extracted from the 100 mL of water containing 600 mg of caffeine with one portion of in 60 mL dichloromethane.

4 0

2 years ago

Read 2 more answers