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mart [117]
3 years ago
15

Solve the problems. Express your answers to the correct number of significant figures.

Chemistry
2 answers:
skad [1K]3 years ago
7 0

Ans: 1) 1.41

        2) 119.3

While adding two numbers, the rule of significant figures requires that the number of decimal places in the final answer is equal to that of the term with the least decimal places

In the given examples:

1) 0.09 - has 2 decimal places

1.324- has 3 decimal places

Sum = 0.09 + 1.324 = 1.414

Round off to 2 decimal places = 1.41

2) 41.0 - has 1 decimal place

78.3- has 1 decimal place

Sum = 41.0 + 78.3 = 119.3

Final answer will also have 1 decimal place = 119.3

Scilla [17]3 years ago
4 0

1.    1.141

2.  119.3

HOPE THIS HELPS.


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A reaction of 41.9 g of Na and 30.3 g of Br2 yields 36.4 g of NaBr . What is the percent yield?
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Answer: The percent yield is, 93.4%

Explanation:

First we have to calculate the moles of Na.

\text{Moles of Na}=\frac{\text{Mass of Na}}{\text{Molar mass of Na}}=\frac{41.9g}{23g/mole}=1.82moles

Now we have to calculate the moles of Br_2

{\text{Moles of}Br_2} = \frac{\text{Mass of }Br_2 }{\text{Molar mass of} Br_2} =\frac{30.3g}{160g/mole}=0.189moles

{\text{Moles of } NaBr} = \frac{\text{Mass of } NaBr }{\text{Molar mass of } NaBr} =\frac{36.4g}{103g/mole}=0.353moles

The balanced chemical reaction is,

2Na(s)+Br_2(g)\rightarrow 2NaBr

As, 1 mole of bromine react with = 2 moles of Sodium

So, 0.189 moles of bromine react with = \frac{2}{1}\times 0.189=0.378 moles of Sodium

Thus bromine is the limiting reagent as it limits the formation of product and Na is the excess reagent.

As, 1 mole of bromine give = 2 moles of Sodium bromide

So, 0.189 moles of bromine give = \frac{2}{1}\times 0.189=0.378 moles of Sodium bromide

Now we have to calculate the percent yield of reaction

\%\text{ yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times 100=\frac{0.353 mol}{0.378}\times 100=93.4\%

Therefore, the percent yield is, 93.4%

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