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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

2 years ago

Can someone answer an easy chemistry question? A chemistry grad student measures the performance of the new pump in his lab. The

ella [17]

This problem requires our calculation to undergo the dimensional analysis approach. In this approach, you disregard the actual quantity and focus on the units of measurement. This helps us know the units of our final answer.

First, let's ignore 16. Let's focus on converting the units kPa-mm³/s to mJ/s. The unit kPa stands for kiloPascals which is 1000 times greater than 1 Pa. The unit mJ, on the other hand, stands for millijoules, which is 1000 times lesser than Joules. The relationship between the two is that, Joules = Pa × m³. But since we want our final answer to be mJ, that would be equal to Pa×mm³. Since the original unit already contains mm³, all we have to do is convert kPa to Pa.

16 kPa-mm³/s * (1000 Pa/1 kPa) = 16,000 Pa-mm³/s

Since Pa-mm³ is equal to mJ, the final conversion yields to 16,000 Pa-mm³/s.

7 0

2 years ago

Read 2 more answers

Please help, I'm confused and it's due very soon!

ryzh [129]

Answer:

1. Lanthanum-139 atom is the stable isotope of lanthanum with relative atomic mass 138.906348, 99.9 atom percent natural abundance and nuclear spin 7/2.

2. In fact, over 80% of electric cars sold globally utilized permanent magnet-based motors in 2019. These magnets are typically made with rare-earth materials such as neodymium and dysprosium, which have a very geographically constrained supply chain.

3. It is the second most reactive of the rare-earth metals after europium. Lanthanum oxidizes in air at room temperature to form La2O3. It slowly reacts with water and quickly dissolves in diluted acids, except hydrofluoric acid (HF) because of formation of a protective fluoride (LaF3) layer on the surface of the metal.

Explanation:

3 0

2 years ago

A sample of a hydrocarbon is found to contain 7.48g carbon and 2.52g hydrogen. What is the empirical

Aliun [14]

Answer:

CH₄

Explanation:

To determine the empirical formula of the hydrocarbon, we need to follow a series of steps.

Step 1: Determine the mass of the compound

The mass of the compound is equal to the sum of the masses of the elements that form it.

m(CxHy) = mC + mH = 7.48 g + 2.52 g = 10.00 g

Step 2: Calculate the percent by mass of each element

%C = mC / mCxHy × 100% = 7.48 g / 10.00 g × 100% = 74.8%

%H = mH / mCxHy × 100% = 2.52 g / 10.00 g × 100% = 25.2%

Step 3: Divide each percentage by the atomic mass of the element

C: 74.8/12.01 = 6.23

H: 25.2/1.01 = 24.95

Step 4: Divide both numbers by the smallest one, i.e. 6.23

C: 6.23/6.23 = 1

H: 24.95/6.23 ≈ 4

The empirical formula of the hydrocarbon is CH₄.

6 0

3 years ago

When propane (C3H8) is burned in air (O2), it completely combusts and produces water (H2O) and carbon dioxide (CO2). How many mo

Damm [24]

Answer:

20 mol

Explanation:

First, balance the chemical reaction

C3H8 + 5O2 --> 4H2O + 3CO2

Using the mole ratio, we can multiply the moles of propane by 4 to find the moles of H2O.

5 × 4= 20

20 mol

6 0

2 years ago