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2 years ago

Which is a function of the gallbladder

Chemistry

1 answer:

Alinara [238K]2 years ago

7 0

Answer:

<h3>Gall Bladder:</h3>

The gallbladder is a pear-shaped, hollow structure located under the liver and on the right side of the abdomen. Its primary function is to store and concentrate bile, a yellow-brown digestive enzyme produced by the liver.

Without a gallbladder, there's no place for bile to collect. Instead, your liver releases bile straight into the small intestine. This allows you to still digest most foods. However, large amounts of fatty, greasy, or high-fiber food become harder to digest.

The gallbladder is part of the biliary tract. The gallbladder serves as a reservoir for bile while it's not being used for digestion. The gallbladder's absorbent lining concentrates the stored bile.

<h2>HOPE U UNDERSTOOD</h2>

THANKS★

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In ____________ reactions, certain molecules are broken down; in the process they release energy which is captured when high-ene

Rudik [331]

Answer:

Exergonic ,Endergonic,low concentration area,high

Explanation:

In exergonic reaction,certain molecules are broken down;in the process they release energy which is captured when high energy molecules(such as ATP and NADH) are formed.

The breakdown of these molecules can be coupled to thermodynamically unfavorable processes such as Endergonic reactions or pumping og hydrogen ion from low concentration areas to high concentration areas.

3 0

2 years ago

Phosphorous trichloride and phosphorous pentachloride equilibrate in the presence of molecular chlorine according to the reactio

umka21 [38]

Answer : The value of $K_p$ at this temperature is 66.7

Explanation : Given,

Pressure of $PCl_3$ at equilibrium = 0.348 atm

Pressure of $Cl_2$ at equilibrium = 0.441 atm

Pressure of $PCl_5$ at equilibrium = 10.24 atm

The balanced equilibrium reaction is,

$PCl_3(g)+Cl_2(g)\rightleftharpoons PCl_5(g)$

The expression of equilibrium constant $K_p$ for the reaction will be:

$K_p=\frac{(p_{PCl_5})}{(p_{PCl_3})(p_{Cl_2})}$

Now put all the values in this expression, we get :

$K_p=\frac{(10.24)}{(0.348)(0.441)}$

$K_p=66.7$

Therefore, the value of $K_p$ at this temperature is 66.7

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2 years ago

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2 years ago

The vapor pressure of water at 25.0°C is 23.8 torr. Determine the mass of glucose (molar mass = 180 g/mol) needed to add to 500.

mrs_skeptik [129]

According to Raoult's law the relative lowering of vapour pressure of a solution made by dissolving non volatile solute is equal to the mole fraction of the non volatile solute dissolved.

the relative lowering of vapour pressure is the ratio of lowering of vapour pressure and vapour pressure of pure solvent

$\frac{p^{0-}p}{p^{0}}=x_{B}$

Where

xB = mole fraction of solute=?

$p^{0}=23.8torr$

p = 22.8 torr

$x_{B}=\frac{23.8-22.8}{23.8}=0.042$

mole fraction is ratio of moles of solute and total moles of solute and solvent

moles of solvent = mass / molar mass = 500 /18 = 27.78 moles

putting the values

$molefraction=\frac{molessolute}{molesolute+molessolvent}$

$0.042=\frac{molessolute}{27.78+molessolute}$

$1.167+0.042(molesolute)=molessolute$

$molessolute=1.218$

mass of glucose = moles X molar mass = 1.218 X 180 = 219.24 grams

3 0

3 years ago

A certain reaction with an activation energy of 185 kJ/mol was run at 505 K and again at 525 K . What is the ratio of f at the h

frosja888 [35]

Answer:

The ratio of f at the higher temperature to f at the lower temperature is 5.356

Explanation:

Given;

activation energy, Ea = 185 kJ/mol = 185,000 J/mol

final temperature, T₂ = 525 K

initial temperature, T₁ = 505 k

Apply Arrhenius equation;

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]$

Where;

$\frac{f_2}{f_1}$ is the ratio of f at the higher temperature to f at the lower temperature

R is gas constant = 8.314 J/mole.K

$Log(\frac{f_2}{f_1} ) = \frac{E_a}{2.303 \times R} [\frac{1}{T_1} -\frac{1}{T_2} ]\\\\Log(\frac{f_2}{f_1} ) = \frac{185,000}{2.303 \times 8.314} [\frac{1}{505} -\frac{1}{525} ]\\\\Log(\frac{f_2}{f_1} ) = 0.7289\\\\\frac{f_2}{f_1} = 10^{0.7289}\\\\\frac{f_2}{f_1} = 5.356$

Therefore, the ratio of f at the higher temperature to f at the lower temperature is 5.356

5 0

3 years ago