What is a liquid mixture whose parts are evenly blended?.

Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange

SIZIF [17.4K]

Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

Explanation:

$CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)$

A

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

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

$100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 10.0 g

10.0 grams of ethyl butyrate would be synthesized.

B

Theoretical yield of ethyl butyrate = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

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

$=\frac{5.75 g}{10.0 g}\times 100=57.5\%$

57.5% was the percent yield.

C

Moles of butanoic acid = $\frac{7.60 g}{88 g/mol}=0.08636 mol$

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

$\frac{1}{1}\times 0.08636 mol=0.08636 mol$ of ethyl butyrate

Mass of 0.08636 moles of ethyl butyrate =

0.08636 mol × 116 g/mol = 10.0 g

Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

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

$78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100$

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

8 0

3 years ago

: An unknown metal crystallizes in the cubic crystal structure. The metal has a radius 140pm, atomic mass of 135 g/mol, and dens

ki77a [65]

The cubic unit cell this metal crystallize as is BCC structure .

The structure of a crystalline solid, whether a metal or not, is best described by considering its simplest repeating unit, which is referred to as its unit cell.

The unit cell consists of lattice points that represent the locations of atoms or ions.

The entire structure then consists of this unit cell repeating in three dimensions

$\rm \rho =\dfrac{nM}{N_{0} a^{3}} \\\\\\\\\rm n =\dfrac{\rho N_{0} a^{3}}{M} \\\\\\\\Assuming\; it \; to \; be \; a\; BCC\; structure \\\\\\ r = \dfrac{\sqrt{3} \times a}{4} \\\\\\Therefore\; a = 3.23 \times 10^{-8}\\\\\\\\\rm n =\dfrac{13.3 \times 6.022 \times 10^{23}\times 3.23^{3}\times (10^{-8})^{3}}{135} \\\\$

n= 2

Hence our assumption was correct

It is a BCC structure .

Therefore the cubic unit cell this metal crystallize as is BCC structure .

To know more about unit cell

brainly.com/question/13110055

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5 0

2 years ago

Balance the equation for the reaction in which sodium oxide reacts with water to form sodium hydroxide.

tensa zangetsu [6.8K]

Answer:

$➢ \: Balance \: the \: equation \: for \: the \: reaction \\ in \: which \: sodium \: oxide \: reacts \\ with \: water \: to \: form \: sodium \: hydroxide.$

⇒ We have Na2O + H2O --> NaOH. We have 2 sodiums and 2 oxygens and 2 hydrogens on the left side, but only one of each on the right side.

Sodium Oxide + Water → Sodium Hydroxide

⇒ Na2O + H2O → 2NaOH .

Sodium oxide is used in ceramics and glasses. Sodium oxide reacts exothermically with cold water to produce sodium hydroxide solution.

6 0

2 years ago

What is the concentration of a solution in which 10.0 g of AgNO, is dissolved in 450 mL of solution?

Rufina [12.5K]

Answer:

M=0.15

Explanation:

138 g AgNO -> 1 mol AgNO

10 g AgNO -> x

x= (10 g AgNO * 1 mol AgNO)/138 g x=0.07 mol AgNO

450 mL=0.45 L

M= mol solute/L solution

M= 0.07 mol AgNO/0.45L

M=0.15

4 0

3 years ago

How many moles are in sample containing 2.71 x 10^24 atoms of iron?

Deffense [45]

Answer:

4.5 moles

Explanation:

One mole is equal to 6.022 x 10^23 atoms

2.71 x 10^24 atoms * 1 mol/ 6.022 x 10^23 atoms = 4.5 moles

5 0

3 years ago