Answer:
in the attached image is the reaction mechanism.
Explanation:
The first reaction (reaction 1) shown in the attached image is the Wolff-Kishner reduction, which is characterized when the carbonyl is reduced to an alkane in the presence of a hydrazine and a base. In reaction 1, the aldehyde reacts with hydrazine to produce oxime. This mechanism begins with the attack of the amine on the carbonyl group. Proton exchange happens and the water leaves the molecule.
In reaction 2, the KOH is deprotoned in nitrogen and organized to form the bond between the nitrogen molecule. this deprotonation releases the nitrogen gas
The equilibrium membrane potential is 41.9 mV.
To calculate the membrane potential, we use the <em>Nernst Equation</em>:
<em>V</em>_Na = (<em>RT</em>)/(<em>zF</em>) ln{[Na]_o/[Na]_ i}
where
• <em>V</em>_Na = the equilibrium membrane potential due to the sodium ions
• <em>R</em> = the universal gas constant [8.314 J·K^(-1)mol^(-1)]
• <em>T</em> = the Kelvin temperature
• <em>z</em> = the charge on the ion (+1)
• <em>F </em>= the Faraday constant [96 485 C·mol^(-1) = 96 485 J·V^(-1)mol^(-1)]
• [Na]_o = the concentration of Na^(+) outside the cell
• [Na]_i = the concentration of Na^(+) inside the cell
∴ <em>V</em>_Na =
[8.314 J·K^(-1)mol^(-1) × 293.15 K]/[1 × 96 485 J·V^(-1)mol^(-1)] ln(142 mM/27 mM) = 0.025 26 V × ln5.26 = 1.66× 25.26 mV = 41.9 mV
6.02*10^23 is Avagadro's number representing the number of molecules per mole of substance.
Answer: The correct answer is D. 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit.
Explanation:
Conversion of degree Celsius to Kelvin :
K=^oC+273
Conversion of degree Celsius to degrees Fahrenheit :
^oF=(\frac{9}{5}\times ^oC)+32
By using these two conversion factors, we get the three temperature readings all mean the same thing.
For option A :
K=^oC+273=100+273=373K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF
For option B :
K=^oC+273=100+273=373K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 100)+32=212^oF
For option C :
K=^oC+273=0+273=273K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF
For option D :
K=^oC+273=0+273=273K
^oF=(\frac{9}{5}\times ^oC)+32=(\frac{9}{5}\times 0)+32=32^oF
From the given options, only option (D) is correct.
Hence, the correct option is, (D) 273 Kelvin, 0 degrees Celsius, 32 degrees Fahrenheit
Hope this helps!
B............................................................................