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
0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.
Potassium hydrogen phthalate, KHP, is a monoprotic acid often used to standardize NaOH solutions.
The balanced neutralization equation is:
NaOH(aq) + KHC₈H₄O₄(aq) ⇒ KNaC₈H₄O₄(aq) + H₂O(l)
- Step 1: Calculate the reacting moles of KHP.
0.212 g of KHP react. The molar mass of KHP is 204.22 g/mol.
0.212 g × 1 mol/204.22 g = 1.04 × 10⁻³ mol
- Step 2: Determine the reacting moles of NaOH.
The molar ratio of NaOH to KHP is 1:1.
1.04 × 10⁻³ mol KHP × 1 mol NaOH/1 mol KHP = 1.04 × 10⁻³ mol NaOH
- Step 3: Calculate the molarity of NaOH.
1.04 × 10⁻³ moles of NaOH are in 35.00 mL of solution.
[NaOH] = 1.04 × 10⁻³ mol / 35.00 × 10⁻³ L = 0.0297 M
0.212 g of KHP is are dissolved in 50.00 mL of water and are titrated by 35.00 mL of 0.0297 M NaOH.
Learn more about titration here: brainly.com/question/4225093
Answer:
54.4 mol
Explanation:
the equation for complete combustion of butane is
2C₄H₁₀ + 13O₂ ---> 8CO₂ + 10H₂O
molar ratio of butane to CO₂ is 2:8
this means that for every 2 mol of butane that reacts with excess oxygen, 8 mol of CO₂ is produced
when 2 mol of C₄H₁₀ reacts - 8 mol of CO₂ is produced
therefore when 13.6 mol of C₄H₁₀ reacts - 8/2 x 13.6 mol = 54.4 mol of CO₂ is produced
therefore 54.4 mol of CO₂ is produced