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Phoenix [80]
3 years ago
14

How many moles are in 150 g of Iron (III) oxide? (Fe203)

Chemistry
2 answers:
artcher [175]3 years ago
8 0

Answer:

The answer is 159.6882

ollegr [7]3 years ago
6 0

Answer:

0.939moles

Explanation:

no. of moles = mass/molar mass

= 150/159.7

= 0.939moles

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What mass of radon is in 8.17 moles of radon?
Len [333]

Answer:

1813.74g

Explanation:

Given parameters:

Number of moles of radon  = 8.17moles

Unknown:

Mass of radon  = ?

Solution:

To solve this problem, we use the expression below:

      Number of moles = \frac{mass}{molar mass}  

Molar mass of radon  = 222g/mol

Now insert the parameters and solve;

    Mass of radon  = Number of moles x molar mass

                              = 8.17 x 222

                              = 1813.74g

4 0
3 years ago
A teacher makes the following statement.
melamori03 [73]
The correct answer among the choices given is the first option.The teacher most likely is talking about distillation of a mixture. Distillation is a unit operation that separates component substances from a liquid mixture which is shown by the teacher. Also, the most common purifying technique in the production of gasoline is by this process.
5 0
2 years ago
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Which of the following are the bark, roots, seeds, buds or berries of an aromatic plant?
djverab [1.8K]

Answer:

Spices

Explanation:

Herbs are the leaves.

Vegetables are the seeds.

Fruits are the seeds.

So spices are the only option left.

8 0
2 years ago
PLEASE HELP ME!!! I WILL GIVE BRAINLIEST!!
attashe74 [19]
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3 years ago
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For the following reaction, 4.07 grams of aluminum oxide are mixed with excess sulfuric acid. The reaction yields 10.4 grams of
torisob [31]

Answer:

Theoretical yield = 13.7 g

% yield =76 %

Explanation:

For Al_2O_3

Mass of Al_2O_3  = 4.07 g

Molar mass of Al_2O_3  = 101.96 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

Moles= \frac{4.07\ g}{101.96\ g/mol}

Moles\ of\ Al_2O_3= 0.0399\ mol

According to the reaction:

Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O

1 mole of Al_2O_3  on reaction produces 1 mole of Al_2(SO_4)_3

So,  

0.0399 mole of Al_2O_3  on reaction produces 0.0399 mole of Al_2(SO_4)_3

Moles of Al_2(SO_4)_3  obtained = 0.0399 mole

Molar mass of Al_2(SO_4)_3 = 342.2 g/mol

The formula for the calculation of moles is shown below:

moles = \frac{Mass\ taken}{Molar\ mass}

Thus,

0.0399= \frac{Mass}{342.2\ g/mol}

Mass= 13.7\ g

<u>Theoretical yield = 13.7 g</u>

The expression for the calculation of the percentage yield for a chemical reaction is shown below as:-

\%\ yield =\frac {Experimental\ yield}{Theoretical\ yield}\times 100

Given , Values from the question:-

Theoretical yield = 13.7 g

Experimental yield = 10.4 g

Applying the values in the above expression as:-

\%\ yield =\frac{10.4}{13.7}\times 100

<u>% yield =76 %</u>

6 0
2 years ago
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