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zhuklara [117]
3 years ago
8

Examine the chemical reaction and lab scenario.

Chemistry
1 answer:
Ierofanga [76]3 years ago
3 0

Explanation:

Percentage Yield

= (3.37g/3.81g) * 100% = 88.45%.

Therefore 88.45% SO2 is the percentage yield.

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How much mass do 3.4 moles of silver contain?
tensa zangetsu [6.8K]

367.2 g of silver

Explanation:

To find the mass of a substance knowing the number of moles we use the following formula:

number of mole = mass / molecular weight

In the case of silver we use the atomic weight of 108 g/mole.

mass = number of moles × molecular weight

mass of silver = 3.4 moles × 108 g/mole

mass of silver = 367.2 g

Learn more about:

moles

brainly.com/question/2293005

#learnwithBrainly

5 0
2 years ago
Different measurements are expressed in different units. Choose the correct Sl units for the lo
swat32

Answer:

The SI units for measuring the velocity of the car: meters per second (m/s)

The Sl units for measuring the acceleration of the car: meters per second squared  (m/s^2)

The SI units for measuring force:  Newton (N)

The SI units for measuring mass: kilograms (kg)

Explanation:

Système international (SI) or International System of Units consist of a list of unit measurement that mostly used by scientist. The scientist from a different country might use different unit that makes them have to convert the result of each other. The usage of the same unit measurement will help the scientist to read that publication easier, make it easier to share and discuss any topic. The unit used is metric since the decimal system also makes the conversion of a unit easier.  

8 0
2 years ago
How many grams of fluorine are needed to generate 7.65 moles Carbon Tetrafluoride
Delicious77 [7]

Answer:

36.3375 g

Explanation:

7 0
2 years ago
15
Len [333]
A.) AlO is the correct formula
8 0
3 years ago
Find percent yield:
saveliy_v [14]

<u>Answer:</u> The percent yield of the reaction is 91.8 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}     .....(1)

  • <u>For B_5H_9 :</u>

Given mass of B_5H_9 = 4.0 g

Molar mass of B_5H_9 = 63.12 g/mol

Putting values in equation 1, we get:

\text{Moles of }B_5H_9=\frac{4g}{63.12g/mol}=0.0634mol

  • <u>For oxygen gas:</u>

Given mass of oxygen gas = 10.0 g

Molar mass of oxygen gas = 32 g/mol

Putting values in equation 1, we get:

\text{Moles of oxygen gas}=\frac{10g}{32g/mol}=0.3125mol

The chemical equation for the reaction of B_5H_9 and oxygen gas follows:

2B_5H_9+12O_2\rightarrow 5B_2O_3+9H_2O

By Stoichiometry of the reaction:

12 moles of oxygen gas reacts with 2 moles of B_2H_5

So, 0.3125 moles of oxygen gas will react with = \frac{2}{12}\times 0.3125=0.052mol of B_2H_5

As, given amount of B_2H_5 is more than the required amount. So, it is considered as an excess reagent.

Thus, oxygen gas is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

12 moles of oxygen gas produces 5 moles of B_2O_3

So, 0.3125 moles of oxygen gas will produce = \frac{5}{12}\times 0.3125=0.130moles of water

Now, calculating the mass of B_2O_3 from equation 1, we get:

Molar mass of B_2O_3 = 69.93 g/mol

Moles of B_2O_3 = 0.130 moles

Putting values in equation 1, we get:

0.130mol=\frac{\text{Mass of }B_2O_3}{69.63g/mol}\\\\\text{Mass of }B_2O_3=(0.130mol\times 69.63g/mol)=9.052g

To calculate the percentage yield of B_2O_3, we use the equation:

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

Experimental yield of B_2O_3 = 8.32 g

Theoretical yield of B_2O_3 = 9.052 g

Putting values in above equation, we get:

\%\text{ yield of }B_2O_3=\frac{8.32g}{9.052g}\times 100\\\\\% \text{yield of }B_2O_3=91.8\%

Hence, the percent yield of the reaction is 91.8 %

6 0
3 years ago
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