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joja [24]
3 years ago
14

Imagine a beaker divided down the center by a rigid membrane that is freely permeable to water but impermeable to glucose. Side

1 contains a 10% glucose solution & side 2 contains the same volume of pure water. At equilibrium, what will be the situation?
1. water will continue to move from side 2 to side 1
2. the volume of liquid remain equal on both sides
3. water will continue to move from side 1 to side 2
4. the volume of liquid will be greater in side 2
5. the volume of liquid will be greater in side 1
Chemistry
1 answer:
damaskus [11]3 years ago
6 0

Answer:

D. The volume of liquid will be greater in side 1

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Answer:

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Explanation:

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2 years ago
Consider two solid blocks, one hot and the other cold, brought into contact in an adiabatic container. After awhile, thermal equ
Alex Ar [27]

Answer: The statement is not correct because the decrease in entropy of the hot solid CANNOT BE equal to the increase in entropy of the cold one

Explanation:

Let us start by stating the second law of thermodynamics and it the second law of thermodynamics states that there is an entity called entropy and entropy will always increase all the time. Also, the second law of thermodynamics states that the change in entropy can never be negative. The second law of thermodynamics can be said to be equal to Change in the transfer of heat, all divided by temperature.

So, the first law of thermodynamics deals with the conservation of energy. But there is nothing like conservation of entropy.

Therefore, the decrease in entropy of the hot solid CANNOT BE equal to the increase in entropy of the cold one because entropy is not a conserved property.

7 0
3 years ago
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Virtually no chemical reaction can occur in the body in the absence of enzymes. how might excessively high body temperature or a
olchik [2.2K]
1. The structural unit of nucleic acids are composed of repeating units of monomers called nucleotides. Nucelotides are composed of three functional groups: sugars which are specifically pentoses (5-Carbon sugars), phosphate group and nitrogenous base.

2. The two major classes of nucleic acids in the body are the DNA or deoxyribonucleic acids and RNA or ribonucleic acids.

3.
a. Based on the nitrogenous bases and sugar, the DNA has a deoxyribose as the sugar and its 4 bases are adenine, guanine, cytosine and thymine. For RNA, the sugar is ribose while its 4 bases are <span>adenine, guanine, cytosine, and uracil

b. Based on the </span>general three-dimensional structure, DNA is a double stranded β-helix with a long chain of nucleotides. RNA is composed of a shorter chain with a single strand α-helix structure.

c. Based on r<span>elative functions, the DNA is responsible for storing the genetic information while the RNA is responsible for transporting the genetic information to the ribosomes which synthesize proteins.</span>
8 0
3 years ago
Sunflower oil contains 0.080 mol palmitic acid (C16H32O2)/mol, 0.060 mol stearic acid (C18H36O2)/mol, 0.27 mol oleic acid (C18H3
Cerrena [4.2K]

Answer:

x_H=0.882

x_N=0.118

Explanation:

In this reactor, oleic and linoleic acid react with hydrogen to form stearic acid. This reactions can be represented by:

Oleic: C_{18}H_{34}O_2 (l) + H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)

Linoleic: C_{18}H_{32}O_2 (l) + 2 H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)

Having this reactions in mind, the first thing is to determine the moles of hydrogen required:

<u>Base of caculation: 1 mol of sunflower oil</u>

For oleic acid: n_{Holeic}=\frac{1 mol H_2}{1 mol oleic}*\frac{0.27 mol oleic}{1 mol oil}*\frac{335 mol oil}{hr}

n_{Holeic}=frac{90.45 mol H_2}{hr}

For linoleic acid: n_{Hlinoleic}=\frac{2 mol H_2}{1 mol linoleic}*\frac{0.59 mol linoleic}{1 mol oil}*\frac{335 mol oil}{hr}

n_{Holeic}=frac{395.3 mol H_2}{hr}

n_{Htotal}=\frac{90.45 mol H_2}{hr}+\frac{395.3 mol H_2}{hr}

n_{Htotal}=frac{485.75 mol H_2}{hr}

Applying the excess:

n_{Htotal}=frac{485.75 mol H_2}{hr}*1.65=801.48 mol

Nitrogen: n_N= 801.48 mol*\frac{0.05 mol N}{0.95 mol}

n_N= 42.2 mol N

<u>After the reactions</u>:

n_H=801.48 mol-485.75mol=315.73 mol

and the nitrogen is inert.

Purge stream:

n_total=42.2+315.73 mol=357.93 mol

x_H=\frac{315.73mol}{357.93mol}=0.882

x_N=\frac{42.2mol}{357.93mol}=0.118

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Answer:C

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3 years ago
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