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Novosadov [1.4K]
3 years ago
13

Secondary bile salts lack a hydroxyl (OH) group when compared to primary bile salts. What is the consequence of the loss of this

group?
Chemistry
1 answer:
Slav-nsk [51]3 years ago
7 0

Answer:

When secondary bile salts (or acids) lack a OH, they are reabsorbed

Explanation:

The lost of a hydroxyl group (OH) in the bile salts structure means less hydro solubility. Thus, they are not eliminated, conversely, the are reabsorbed for intestine and they go back to the liver.

I hope my answer helps

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What is the pH of an acidic solution?
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The answer is A, between 0 and 7.

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A sample of neon has a volume of 40.81 m3 at 23.5C. At what temperature, in Kelvins, would the gas occupy 50.00 cubic meters? As
mezya [45]

At  \fbox{\begin \\363 K \end{minispace}}  temperature, a sample of neon gas will occupy 50.00 \text{ m}^{3} volume.

Further Explanation:

The given problem is based on the concept of Charles’ law. Charles’ law states that “at constant pressure and fixed mass the volume occupied an ideal gas is directly proportional to the Kelvin temperature.”

Mathematically the law can be expressed as,

\fbox{ \begin \\ V \propto T \end{minispace}}

Or,

\frac{V}{T}=k

Here, <em>V</em> is the volume of the gas, <em>T</em> is Kelvin temperature, and <em>k</em> is proportionality constant.

Given information:

The initial volume of neon gas is 40.81 \text{ m}^{3} .

The final volume of neon gas is  50.00 \text{ m}^{3}.

The initial temperature value is 23.5 \text{ } ^{\circ} \text{C} .

To calculate:

The final temperature

Given Condition:

  • The pressure is constant.
  • Mass of gas is fixed.

Solution:

Step 1: Modify the mathematical expression for Charles’ law for two different temperature and volume values as follows:

\frac{V_{1}}{T_{1}}=\frac{V_{2}}{T_{2}}

Here,

  • V_{1}is the initial volume of the gas.
  • V_{2} is the final volume of the gas.
  • T_{1} is the initial temperature of the gas.
  • T_{2} is the final temperature of the gas.

Step 2: Rearrange equation (2) for .

\fbox {\begin \\T_{2}=\frac{(V_{2}) \times (T_{1})}{V_{1}}\\\end{minispace}}                                                                  …… (2)

Step 3: Convert the given temperature  from degree Celsius to Kelvin.

The conversion factor to convert degree Celsius to Kelvin is,

T(\text{K}) = T(^{\circ}\text{C}) + 273.15                                      …… (3)

Substitute 23.5\text{ }^{\circ} \text{C} for T(^{\circ}\text{C})  in equation (3) to convert temperature from degree Celsius to Kelvin.

T(\text{K}) = 23.5 \text{ } ^{\circ} \text{C} + 273.15\\T(\text{K})= 296.65 \text{ K}

Step 4: Substitute 40.81 \text{ m}^{3}  for V_{1} ,  50.00 \text{ m}^{3} for V_{2}  and  296.65 \text{ K} for T_{1}  in equation (2) and calculate the value of T_{2} .

T_{2}=\frac{(50.00 \text{ m}^{3}) \times (296.65 \text{ K})}{40.81 \text{ m}^{3}}\\T_{2}=363.45 \text{ K}\\T_{2} \approx 363 \text{ K}

Important note:

  • The temperature must be in Kelvin.
  • The condition of fixed mass and fixed pressure must be fulfilled in order to apply Charles’ law.

Learn More:

1. Gas laws brainly.com/question/1403211

2. Application of Charles’ law brainly.com/question/7434588

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: States of matter

Keywords: neon, volume, occupies, temperature, Kelvin, degree Celsius, Charle’s law, constant pressure, fixed mass, 40.81 m^3 , 50.00 m^3 , 23.5 degree C , celsius , 363 K , sates of matter, initial volume, final volume, initial temperature, final temperature, V1 , V2 , T1 , T2 .

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