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tiny-mole [99]
3 years ago
11

compound of aspartame is a dipeptide that is often used as a sugar substitute which functional groups are present

Chemistry
1 answer:
Scrat [10]3 years ago
5 0

Answer:

Carboxyl, primary amine, amide, ester, and phenyl.

Explanation:

The functional groups present in the compound of aspartame are carboxyl, primary amine, amide, ester, and phenyl. Aspartame is an artificial non-saccharide sweetener which is 200 times sweeter than sucrose. This aspartame is commonly used as a sugar substitute in many foods and beverages. It has the trade names such as NutraSweet, Equal, and Canderel.

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Explain how materials are suited for different uses based on their physical and chemical properties?
Leya [2.2K]
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4 0
2 years ago
When air pressure is high what is the weather like​
Basile [38]

Answer:

high-pressure systems normally associate with dry weather and mostly clear skies. This usually brings some light winds of cool, dry air, and brings fair weather.

5 0
3 years ago
The number of grams per mole of something is known as?​
Whitepunk [10]

Answer:

Judging from the wording of he question, you mean units. If that is indeed the case, the answer is g/Mol (grams per mol)

Let me know if my interpretation is incorrect and please tell me what you are actually trying to find.

7 0
3 years ago
Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,
Serhud [2]

The question is incomplete. Her eis the complete question.

Steam reforming methane  (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: K_{c} = 2.10^{-2}

Explanation: The reaction for steam reforming methane is:

CH_{4} + H_{2}O ⇒ CO_{} + 3H_{2}

To calculate the concentration equilibrium constant, first calculate the molarity (\frac{mol}{L}) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

<u>Molarity of CH4</u>:

CH4 = \frac{20}{125} = 0.16M

<u>Molarity of H20</u>:

H2O = \frac{10}{125} = 0.08M

At final temperature:

<u>Molarity of H2</u>:

H2 = \frac{18}{125} = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

<u>Molarity of CO</u>:

CO = \frac{0.144}{3} = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

<u>Molarity of CH4</u>:

CH4 = 0.16 - 0.048 = 0.112M

<u>Molarity of H2O</u>:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }

K_{c} = \frac{0.048.0.144^{3} }{0.112.0.032}

K_{c} = 2.10^{-2}

The concentration equilibrium constant for the process is K_{c} = 2.10^{-2}.

4 0
3 years ago
Which of the following is the correct set up AND answer to convert 6.25 x
BlackZzzverrR [31]

Answer:

Explanation:

How many mols do you have?

1 mol = 6.02 * 10^23 atoms

x mol = 6.25 * 10 ^32 atoms

1/x = 6.02*10^23 / 6.25 * 10^32        Cross multiply

6.02 * 10^23 * x = 1 * 6.25 * 10^32    Divide by 6.02 * 10^23

x = 6.25 * 10*32/ 6.02 ^10^23

x = 1.038 * 10^9 mols which is quite large.

Find the number of grams. (Use the value for copper on your periodic table. I will just use an approximate number.)\

1 mol of copper = 63 grams.

1.038 * 10^9 mols of copper = x

1/1.038 * 10^9 = 63/x         Cross multiply

x = 1.038 * 10^9 * 63

x = 6.54 * 10^10 grams of copper.

7 0
3 years ago
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