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Fittoniya [83]
2 years ago
9

If you start with 75 mg of anisole and excess bromine, and the bromination occurs 3 times, then what is your theoretical yield (

in mg) of tribromoanisole?
Chemistry
1 answer:
ser-zykov [4K]2 years ago
3 0

Answer:

239.6 mg is the  theoretical yield of tribromoanisole.

Explanation:

C_6H_5-OCH_3+3Br_2\rightarrow C_6H_2Br_3-OCH_3+3HBr

Mass of anisole C_6H_5-OCH_3 = 75 mg = 0.075 g

1 mg = 0.001 g

Moles of anisole = \frac{0.075 g}{108 g/mol}=0.0006944 mol

According to reaction, 1 mole anisole gives 1 mole of tribromoanisole.

Then 0.0006944 mole of anisole will give:

\frac{1}{1}\times 0.0006944 mol=0.0006944 mol of tribromoanisole

Mass of 0.0006944 mole of tribromoanisole:

0.0006944 mol × 345 g/mol = 0.2396 g = 239.6 mg

239.6 mg is the  theoretical yield of tribromoanisole.

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How many grams of precipitate will be formed when 20.5 mL of 0.800 M
Anton [14]

Answer:

There will be formed 1.84 grams of precipitate (NaNO3)

Explanation:

<u>Step 1</u>: The balanced equation

CO(NO3)2 (aq) + 2 NaOH (aq) → CO(OH)2 (s) + 2 NaNO3 (aq)

<u>Step 2:</u> Data given

Volume of 0.800 M  CO(NO3)2 = 20.5 mL = 0.0205 L

Volume of 0.800 M NaOH = 27.0 mL = 0.027 L

Molar mass of NaNO3 = 84.99 g/mol

<u>Step 3:</u> Calculate moles of CO(NO3)2

Moles CO(NO3)2  = Molarity * volume

Moles CO(NO3)2  = 0.800 M * 0.0205

Moles CO(NO3)2 = 0.0164 moles

Step 4: Calculate moles NaOH

moles of NaOH = 0.800 M * 0.027 L

moles NaOH = 0.0216 moles

Step 5: Calculate limiting reactant

For 1 mole CO(NO3)2 consumed, we need 2 moles of NaOH to produce 1 mole of CO(OH)2 and 2 moles of NaNO3

NaOH is the limiting reactant. It will completely be consumed.

CO(NO3)2 is in excess. There willbe 0.0216 / 2 = 0.0108 moles of CO(NO3)2 consumed. There will remain 0.0164 - 0.0108 = 0.0056 moles of CO(OH)2

Step  6: Calculate moles of NaNO3

For 2 moles of NaOH consumed, we have 2 moles of NaNO3

For 0.0216 moles of NaOH, we have 0.0216 moles of NaNO3

Step 7: Calculate mass of NaNO3

mass of NaNO3 = moles of NaNO3 * Molar mass of NaNO3

mass of NaNO3 = 0.0216 moles * 84.99 g/mol = 1.84 grams

There will be formed 1.84 grams of precipitate (NaNO3)

5 0
3 years ago
True or false a solid has a constant shape and volume
dusya [7]
The answer you are looking for is True
8 0
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What is the correct term for blocked sunlight and reduced solar radiation as a result of a volcanic eruption?
Anvisha [2.4K]
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How many grams in 6.20 x 10^25 atoms of bromine (Br) ? image attached , will give brainliest
Deffense [45]

Answer:

8239.2g

Explanation:

Given parameters:

Number of atoms in Br  = 6.2 x 10²⁵atoms

Unknown:

Mass of Br = ?

Solution:

From mole concepts, we know that:

       1 mole of a substance contains 6.02 x 10²³ atoms/mol

 Molar mass of Br  = 80g/mol

6.2 x 10²⁵atoms  x \frac{1}{6.02 x 10^{23} } \frac{mol}{atoms} x  80 x \frac{g}{moles}  

          = 8239.2g

8 0
3 years ago
When combing one's hair, the comb becomes positively charged. What happens to the hair being combed?
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Since the comb is positively charged, the hair will stick to the comb.

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