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k0ka [10]
3 years ago
5

Find ∆G for a cell whose potential is +0.24 V with 4 moles of electrons exchanged.

Chemistry
2 answers:
Serga [27]3 years ago
8 0

Step 1.  Use  ∆G = -nFEo.

Step 2.  Plug in n = 4, F = 96,500 and E = +0.24

Step 3.   ∆G = -4 × 96,500 × 0.24    =   -92,640 J   =   -92.64 kJ.

Setler [38]3 years ago
6 0
We need to use the following formula
ΔG= -nF E_{cell}

E_{cell}= 0.24V
n= 4 moles
F= constant= 96500C/mol

let's plug in the values.

ΔG= -(4)(96500)(0.24)= -92640 J or -92.6 kJ
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3 years ago
g A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is __________ compared to a 300 mOsM sol
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Answer:

A solution contains 100mM NaCl, 20mM CaCl2, and 20mM urea. We would say this solution is hypotonic compared to a 300 mOsM solution and hypotonic compared to a cell with 300 mOsM (non-penetrating solutes) interior.

Explanation:

The osmolarity is calculated from the molar concentration of the active particles in the solution. We have a solution that is composed of NaCl, CaCl₂ and urea.

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Compared with 300 mOsm, 280 mOsm has a lower osmolarity, so it is a hypotonic solution.

To compare with a cell's osmolarity, we have to consider only the non-penetrating solutes. Urea is considered a penetrating solute for mammalian cells. So, the osmolarity of non-penetrating solutes (NaCl  and CaCl₂) is calculated as:

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Compared to 300 mOsm solution ⇒ 280 mOsm solution is a hypotonic solution

Compared to a cell with 300 mOsm ⇒ 260 mOsm solution is hypotonic

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