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Svetach [21]
3 years ago
11

Why did they focus on higher alcohols to add to or substitute gasoline instead of ethanol? Check all answers that apply. Select

one or more: a. It is easier to produce higher alcohols in microbes than it is to produce ethanol. b. Higher alcohols have a higher hygroscopicity than ethanol. c. Higher alcohols have a lower vapor pressure than ethanol. d. Higher alcohols have a higher energy density than ethanol.
Chemistry
1 answer:
Varvara68 [4.7K]3 years ago
5 0

Answer:

Because of statements (a), (c) and (d).

Explanation:

Let's evaluate each statement to know why they focus on higher alcohols instead of ethanol:

(a) It is easier to produce higher alcohols in microbes than it is to produce ethanol<u>.</u> This is true<u> since ethanol is the major biofuel in the world because it can be easily produced by fermentation technology developed a long time ago. Recently, higher alcohols are produced from microorganisms that are used as microbial cell factories.</u>                    

(b) Higher alcohols have a higher hygroscopicity than ethanol. This is false since ethanol is more hygroscopic than higher alcohols.

(c) Higher alcohols have a lower vapor pressure than ethanol. This is true, ethanol has a higher vapor pressure than higher alcohols. The vapor pressure is important since it can affect the proper cold starting of the engine.  

(d) Higher alcohols have a higher energy density than ethanol. This is true since the production of higher alcohols as biofuels is more desirable than the ethanol because higher alcohols have a high energy density and other more advantages than the use of ethanol.        

Therefore the answer of why did they focus on higher alcohols to add to or substitute gasoline instead of ethanol is because of the statements  (a), (c) and (d).

I hope it helps you!

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A 36.165 mg36.165 mg sample of a chemical known to contain only carbon, hydrogen, sulfur, and oxygen is put into a combustion an
erma4kov [3.2K]

Answer:

The empirical formula for the compound is C6H12SO2.

Explanation:

We'll begin by writing out what was given from the question. This is shown:

Let us consider the First experiment:

Mass of the compound = 36.165 mg

Mass of CO2 = 64.425 mg

Mass of H2O = 26.373 mg

Data obtained from the Second experiment:

Mass of compound = 47.029 mg

Mass of SO2 = 20.32 mg

Next, we'll determine the mass of C, H and S. This is illustrated below:

Molar Mass of CO2 = 12 + (2x16) = 44g/mol

Mass of C in CO2 = 12/44 x 64.425 Mass of C = 17.57 mg

Molar Mass of H2O = (2x1) + 16 = 18g/mol

Mass of H in H2O = 2/18 x 26.373

Mass of H = 2.93 mg

Molar Mass of SO2 = 32 + (16x2) = 64g/mol

Mass of S in SO2 = 32/64 x 20.32

Mass of S = 10.16 mg

At this stage, it is important we determine the percentage composition of C, H, S and O. This is illustrated below:

% of C = 17.57/36.165 x 100 = 48.58%

% of H = 2.93/36.165 x 100 = 8.10%

% of S = 10.16/47.029 x 100 = 21.60%

% of O = 100 - (48.58 + 8.1 + 21.6)

% of O = 21.72%

Now we can easily obtain the empirical formula for the compound by doing the following.

Step 1:

Divide by their molar mass

C = 48.58/12 = 4.0483

H = 8.10/1 = 8.1

S = 21.60/32 = 0.675

O = 21.72/16 = 1.3575

Step 2:

Divide by the smallest:

C = 4.0483/0.675 = 6

H = 8.1/0.675 = 12

S = 0.675/0.675 = 1

O = 1.3575/0.675 = 2

From the calculations made above, empirical formula for the compound is C6H12SO2

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3 years ago
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Answer:

The answer is 60W

Power = Work done/ time

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Work done = 60J

Power = 60/1

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Hope this helps.

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3 years ago
How many moles are in 20 grams of argon​
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Answer:

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Explanation:

40 g of argon = 1mole

Then 20g of argon is,

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I’m just doing this because it told me to in order to get the answer for my homework?
andrew-mc [135]
Ok good luck on your homework then
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