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Morgarella [4.7K]
3 years ago
12

Which is a function of the gallbladder

Chemistry
1 answer:
Alinara [238K]3 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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Amphetamine (C9H13N)(C9H13N) is a weak base with a pKbpKb of 4.2. You may want to reference (Pages 710 - 713) Section 16.8 while
Soloha48 [4]

Answer:

pH = 10.38

Explanation:

  • C9H13N ↔ C9H20O3N+  +  OH-

∴ molar mass C9H13N = 135.21 g/mol

∴ pKb = - log Kb = 4.2

⇒ Kb = 6.309 E-5 = [OH-][C9H20O3N+] / [C9H13N]

∴ <em>C</em> sln = (205 mg/L )*(g/1000 mg)*(mol/135.21 g) = 1.516 E-3 M

mass balance:

⇒ <em>C</em> sln = 1.516 E-3 = [C9H20O3N+] + [C9H13N]......(1)

charge balance:

⇒ [C9H20O3N+] + [H3O+] = [OH-]; [H3O+] is neglected, come from water

⇒ [C9H20O3N+] = [OH-].......(2)

(2) in (1):

⇒ [C9H13N] = 1.516 E-3 - [OH-]

replacing in Kb:

⇒ Kb = 6.3096 E-5 = [OH-]² / (1.516 E-3 - [OH-])

⇒ [OH-]² + 6.3096 E-5[OH] - 7.26613 E-8 = 0

⇒ [OH-] = 2.3985 E-4 M

∴ pOH = - Log [OH-]

⇒ pOH = 3.62

⇒ pH = 14 - pOH = 14 - 3.62 = 10.38

5 0
3 years ago
Which substance has a coefficient of 2 when you balance the equation for this redox reaction? cu(s) + hno3(aq) → cu(no3)2(aq) +
Mama L [17]
In this redox reaction, the Cu goes from oxidation state of (0) to (+2), therefore it oxidises. N in HNO₃ goes from oxidation state of (+5) to N in NO with oxidation state of (+2) and becomes reduced. 
Cu acts as the reducing reagent and HNO₃ is the oxidising agent.
oxidation half reaction
Cu ---> Cu²⁺ + 2e  --1)
reduction half reaction 
4H⁺ + 3e + NO₃⁻ ---> NO + 2H₂O --2)
to balance the number of electrons , 1) x3 and 2) x2
3Cu ---> 3Cu²⁺ + 6e
8H⁺ + 6e + 2NO₃⁻ ---> 2NO + 4H₂O 
add the 2 equations 
3Cu + 8H⁺ + 2NO₃⁻ --->  3Cu²⁺ + 2NO + 4H₂O 
add 6 nitrate ions to both sides to add up to 8 and form acid with 8H⁺ ions
3Cu + 8HNO₃ --->  3Cu(NO₃)₂ + 2NO + 4H₂O 
Balanced equation for the redox reaction is as follows;
3Cu(s) + 8HNO₃(aq) → 3Cu(NO₃)₂(aq) + 2NO(g) + 4H₂O<span>(l)

NO has a coefficient of 2

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8 0
4 years ago
Please help i really need it!
svp [43]

Answer:

mining away

Explanation:

8 0
3 years ago
The equilibrium constant for the reaction
FinnZ [79.3K]

Answer: The concentrations of Cl_2 at equilibrium is 0.023 M

Explanation:

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Volume of solution = 1 L

Initial concentration of Cl_2 = \frac{0.14mol}{1L}=0.14M

The given balanced equilibrium reaction is,

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Initial conc.           0.14 M           0 M       0M    

At eqm. conc.     (0.14-x) M        (x) M        (x) M

The expression for equilibrium constant for this reaction will be,

K_c=\frac{[CO]\times [Cl_2]}{[COCl_2]}

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

4.63\times 10^{-3}=\frac{x)^2}{(0.14-x)}

By solving the term 'x', we get :

x = 0.023 M

Thus, the concentrations of Cl_2 at equilibrium is 0.023 M

7 0
3 years ago
A 7.0 L sample of gas begins at 2.5 atm and 320. K. What is the new pressure if the temperature is changed to 273 K and the volu
enyata [817]

Answer:

2.1 atm

Explanation:

We are given the following variables to work with:

Initial pressure (P1): 2.5 atm

Initial temperature (T1): 320 K

Final temperature (T2): 273 K

Constant volume: 7.0 L

We are asked to find the final pressure (P2). Since volume is constant, we want to choose a gas law equation that relates initial pressure and temperature to final pressure and temperature. Gay-Lussac's law does this:

\frac{P_{1}}{T_1} =\frac{P_{2}}{T_2} \\

We can rearrange the law algebraically to solve for P_{2}.

{P_{2}} =\frac{(T_2)(P_{1} )}{T_1} \\

Substitute your known variables and solve:

P_2 = \frac{273K(2.5atm)}{320K}  = 2.1 atm

4 0
3 years ago
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