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3 years ago

I don’t know what are the ones that are left ,can someone help me please

Chemistry

1 answer:

disa [49]3 years ago

4 0

think of like a experamint and take a example of a project

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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?

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%

3 0

3 years ago

Calculate the entropy change for the reaction: Fe2O3(s) +3C(s) -> 2Fe(s) + 3CO(g)Entropy data:Fe2O3(s): 90 J/K molC(s): 5.7 J

Annette [7]

Explanation:

We are given: entropy of Fe2O3 = 90J/K.mol

: entropy of C = 5.7J/K.mol

: entropy of Fe = 27.2J/K.mol

: entropy of CO = 198J/K.mol

$\begin{gathered} \Delta S\text{ = S}_{products}-S_{reactants} \\ \\ \text{ = \lparen3}\times198+2\times27.2)-(3\times5.7+90) \\ \\ \text{ = 541.3J/K.mol} \end{gathered}$

Answer:

The correct answer is C.

5 0

1 year ago

Simple question plzz help

PSYCHO15rus [73]

Answer:

Explanation:

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3 years ago

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What is the mole fraction of a nonelectrolyte in an aqueous solution above which the vapor pressure of water is 745 mm Hg at 100

12345 [234]

The mole fraction of a gaseous compound is equal to the ratio of the vapor pressure of the compound to the total pressure of the vessel. In this case, teh vapor pressure of tehe non-electrolyte is 760 - 745 = 15 mmHg. hence the mole fraction is 15 mm Hg / 760 mmHg equal to 0.0197

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Classify CH3CH2NHCH3 as a strong base or a weak base

harina [27]

CH3CH2NHCH3 is a weak base.

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