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Veseljchak [2.6K]
2 years ago
8

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

Chemistry
1 answer:
Sergeu [11.5K]2 years ago
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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Which of these is a benefit of nuclear fusion?
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B. There is a lower risk of runaway chain reactions.

Explanation:

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

a) Find the <em>length in millimetres</em>

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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]

<u>Answer:</u> The pH of acid solution is 4.58

<u>Explanation:</u>

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)

  • <u>For KOH:</u>

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

  • <u>For propanoic acid:</u>

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

<u>Initial:</u>          0.1372         0.04514  

<u>Final:</u>           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
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