3 years ago

What is the formula for Disulfur Pentafluoride?

Chemistry

1 answer:

dolphi86 [110]3 years ago

7 0

Answer:

please mark me as brainliest so difficulty to find me and edit

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Now let us refine our model by noting that there is a second source of Na+ ions in the cell: NaI. Suppose the outside of the cel

MArishka [77]

Answer:

$E=55mV$

Explanation:

Hello,

Considering the given information, the new concentration of Na+inside will be (13.6 + 4) mM = 17.6 mM, and outside will be (150 + 0.04) mM = 150.04 mM due to the previous specified conditions.

Now, the recalculation of the Nerst potential at the supposed temperature of 25 °C (which is modifiable) is done via:

$E=\frac{RT}{zF} ln(\frac{C_{Na^+,outside}}{C_{Na^+,outside}} )\\\\E=\frac{8.314\frac{J}{mol*K}*298K}{1*9.65x10^4\frac{C}{mol} } *ln(\frac{150.04}{17.6} )=0.055V*\frac{1x10^3mV}{1V} \\E=55mV$

Best regards.

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3 years ago