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3 years ago

A submicroscopic particle suspended in a solution has a volume 1.3um^3. What is this volume in liters?

2 answers:

5 0

1000 liters = 1 m^3

and
1.3um^3 = 1.3x10^(-6)m^3

multiply both sides of the first equation by 1.3x10^(-6)
1.3x10^(-6) * 1000 liters = 1.3x10^(-6) * 1m^3
1.3x10^(-3) liters = 1.3x10^(-6) m^3

MariettaO [177]3 years ago

3 0

Answer: $1.3\times 10^{-15} L$ .

Explanation:

Volume of the gas is defined as the space occupied by a substance. It is expressed in units like $cm^3$ , $m^3$ , L and ml.

All these units of volume are inter convertible.

We are given:

Volume of the gas = $1.3\mu m^3$

Converting this unit of volume into 'L' by using conversion factor:

$10^{18}\mu m^3=1m^3$

$1.3\mu m^3=\frac{1}{10^{18}}\times 1.3=1.3\times10^{-18}m^3$

$1m^3=1000L$

Thus $1.3\times 10^{-18}m^3=\frac{1000}{1}\times 1.3\times 10^{-18}=1.3\times 10^{-15} L$

Thus the volume in Liters would be $1.3\times 10^{-15} L$ .

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2 years ago

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?

Licemer1 [7]

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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3 years ago

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Answer:

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Explanation:

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Answer:

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