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Pavel [41]
3 years ago
7

What is the ka for a 0.25 m solution of a monoprotic weak acid with a ph of 4.65

Chemistry
1 answer:
34kurt3 years ago
8 0

Answer:

ka = 2x10⁻⁹

Explanation:

Using the pH we can <u>calculate the molar concentration of H⁺</u>, [H⁺]

  • pH = -log[H⁺] = 4.65
  • 10^{-4.65} =  [H⁺] = 2.24 x 10⁻⁵ M

Then we use the expression of Ka for the equilibrium of a weak monoprotic acid:

HA ↔ H⁺ + A⁻

Where:

ka = \frac{[H^+][A^-]}{[HA]}

ka =  \frac{(2.24*10^{-5})(2.24*10^{-5})}{0.25-2.24*10^{-5}} =  2x10⁻⁹

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By which mechanism would a steroid molecule diffuse into the cell?
LUCKY_DIMON [66]

Answer: Directly through the phospholipid membrane

Explanation:

The cell membrane consist of a phospholipid bilayers structure. In the interior of the membrane, the phospholipid tail are hydrophobic, which makes the cell membrane to be selectively permeable, it is permeable to non polar molecules and impermeable to polar molecules.

Because Steroids are fat soluble, non polar compounds, they can diffuse directly through the hydrophobic, non polar core of the phospholipid bilayer without the use of carrier proteins.

5 0
3 years ago
The standard cell potential Ec for the reduction of silver ions with elemental copper is 0.46V at 25 degrees celsius. calculate
Cloud [144]

Answer : The \Delta G for this reaction is, -88780 J/mole.

Solution :

The balanced cell reaction will be,  

Cu(s)+2Ag^+(aq)\rightarrow Cu^{2+}(aq)+2Ag(s)

Here, magnesium (Cu) undergoes oxidation by loss of electrons, thus act as anode. silver (Ag) undergoes reduction by gain of electrons and thus act as cathode.

The half oxidation-reduction reaction will be :

Oxidation : Cu\rightarrow Cu^{2+}+2e^-

Reduction : 2Ag^++2e^-\rightarrow 2Ag

Now we have to calculate the Gibbs free energy.

Formula used :

\Delta G^o=-nFE^o

where,

\Delta G^o = Gibbs free energy = ?

n = number of electrons to balance the reaction = 2

F = Faraday constant = 96500 C/mole

E^o = standard e.m.f of cell = 0.46 V

Now put all the given values in this formula, we get the Gibbs free energy.

\Delta G^o=-(2\times 96500\times 0.46)=-88780J/mole

Therefore, the \Delta G for this reaction is, -88780 J/mole.

7 0
3 years ago
Help me answer this question please?
Tema [17]

Answer:

I think D

Explanation:

Ok, I'm not sure but it sounds right ish you should check a practice video or something. It might also be B or C but im pretty certain it isnt A just ask yourself is the student measuring it in newtons? Is that important in the process? What about if the student is considering the affect of mass is it important? Good luck srry if im not much of help! If this is like A SUPER IMPORTANT TEST OR SOMETHING RLLLLLLLY IMPORTANT just wait for another answer gl!

5 0
3 years ago
Instead of developing embryo inside the egg, the embryo grows and develops inside the mother's womb until the baby is born. This
ra1l [238]

Answer:

mammal

Explanation:

mammals are the types of animals who dont have eggs

8 0
3 years ago
A tank of gas is found to exert 8.6 atm at 38°C. What would be the required
Vesna [10]

Answer:

36.2 K

Explanation:

Step 1: Given data

  • Initial pressure of the gas (P₁): 8.6 atm
  • Initial temperature of the gas (T₁): 38°C
  • Final pressure of the gas (P₂): 1.0 atm (standard pressure)
  • Final temperature of the gas (T₂): ?

Step 2: Convert T₁ to Kelvin

We will use the following expression.

K = °C +273.15

K = 38 °C +273.15 = 311 K

Step 3: Calculate T₂

We will use Gay Lussac's law.

P₁/T₁ = P₂/T₂

T₂ = P₂ × T₁/P₁

T₂ = 1.0 atm × 311 K/8.6 atm = 36.2 K

6 0
3 years ago
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