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3 years ago

What is the mass of ethyl alcohol that fills a 250.0-mL graduated cylinder? The density of ethyl alcohol is 0.789 g/mL.

Chemistry

2 answers:

solong [7]3 years ago

4 0

Answer:

$\boxed {\tt 197.25 \ grams }$

Explanation:

The density formula is:

$d=\frac{m}{v}$

Rearrange the formula for m, the mass. Multiply both sides of the equation by v.

$d*v=\frac{m}{v} *v$

$d*v=m$

Mass can be found by multiplying the density and volume. The density of ethyl alcohol is 0.789 grams per milliliter. The alcohol fills a 250 milliliter graduated cylinder, so the volume is 250 milliliters.

$d= 0.789 \ g/mL \\v= 250 mL$

Substitute the values into the formula.

$0.789 \ g/mL * 250 \ mL=m$

Multiply. Note that the milliliters, or mL will cancel each other out.

$0.789 \ g * 250=m$

$197.25 \ g=m$

The mass of the ethyl alcohol is 197.25 grams.

UNO [17]3 years ago

3 0

Answer: 157.8 g

Explanation: I hope this help my child UwU

What is the weight of the ethyl alcohol that exactly fills a 200.0 mL container the density of ethyl alcohol is 0.789 g ml?

Density review

What is the weight of the ethyl alcohol that exactly fills a 200.0 mL container? The density of ethyl alcohol is 0.789 g/mL.

g = (0.789 g/mL) (200.0 mL) = 158 g

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Vilka [71]

<u>Answer:</u> The amount of hex-3-ene used is 0.0711 moles and the percent yield of 3-hexanol is 59.56 %

<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)

Given mass of hex-3-ene = 5.84 g

Molar mass of hex-3-ene = 82.14 g/mol

Putting values in equation 1, we get:

$\text{Moles of hex-3-ene}=\frac{5.84g}{82.14/mol}=0.0711mol$

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$\text{hex-3-ene}+H_2O\xrightarrow []{10\% H_2SO_4} \text{3-hexanol}$

By Stoichiometry of the reaction:

1 mole of hex-3-ene produces 1 mole of 3-hexanol

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Molar mass of 3-hexanol = 102.2 g/mol

Moles of 3-hexanol = 0.0711 moles

Putting values in equation 1, we get:

$0.0711mol=\frac{\text{Mass of 3-hexanol}}{102.2g/mol}\\\\\text{Mass of 3-hexanol}=(0.0711mol\times 102.2g/mol)=7.27g$

To calculate the percentage yield of 3-hexanol, we use the equation:

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

Experimental yield of 3-hexanol = 4.33 g

Theoretical yield of 3-hexanol = 7.27 g

Putting values in above equation, we get:

$\%\text{ yield of 3-hexanol}=\frac{4.33g}{7.27g}\times 100\\\\\% \text{yield of 3-hexanol}=59.56\%$

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