Hello!
Percentage error = (true value - measured value)/true value x 100%
Percentage error = (27.7-27.0)/27.7 x100% = 25.3%
Answer:
The reaction shifts to the left.
Explanation:
Equilibrium constant (K) = 46.3
Reaction Quotient (Q) = 525
The relationship between Q and K with their implications are given as;
K = Q (No net reaction)
K > Q (Reaction shifts to the right)
K < Q (Reaction shifts to the left)
Since in this question, Q (525) > K (46.3)
The reaction shifts to the left.
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
Answer:
6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.
Explanation:
We are given the chemical equation:
And we want to determine the amount of products produced when 12.5 moles of NH₃ is reacted with excess CuO.
Compute using stoichiometry. From the equation, we can see the following stoichiometric ratios:
- The ratio between NH₃ and N₂ is 2:1. (i.e. One mole of N₂ is produced from every two moles of NH₃.)
- The ratio between NH₃ and Cu is 2:3.
- The ratio between NH₃ and H₂O is 2:3. (i.e. Three moles of H₂O or Cu is produced frome every two moles of NH₃.)
Dimensional Analysis:
- The amount of N₂ produced:
- The amount of Cu produced:
- And the amount of H₂O produced:
In conclusion, 6.25 moles of N₂ is produced, and 18.8 moles of Cu and H₂O is produced.
The class can break or if you put it in cold water it can cool down down fast.
