For a voltaic cell consisting of chromium, an electrode dipped in a 1.20 M chromium (III) nitrate solution and a tin electrode dipped in a 0.400 M tin (II) nitrate solution, the cell potential at 298 K is mathematically given as
Ecell = 0.577 V
<h3 /><h3>What is the cell potential at 298 K?</h3>
Generally, the equation for the Oxidation and Reduction is mathematically given as
Cr(s) ------------------ Cr+3(aq) + 3e- ] x 2 ...O
Sn+2(aq) + 2e- ------------ Sn(s) ] x 3 ...R
Reaction
2 Cr(s) + 3 Sn+2(aq) --------------- 2 Cr+3(aq) + 3 Sn(s)
Therefore
Eicell = - 0.14 - ( - 0.74)
Eicell = 0.60
In conclusion
![Ecell= E0cell - \frac{0.0591}{n} * \frac{log[Cr+3]^2}{ [ Sn+2]^3}](https://tex.z-dn.net/?f=Ecell%3D%20E0cell%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20%2A%20%5Cfrac%7Blog%5BCr%2B3%5D%5E2%7D%7B%20%5B%20Sn%2B2%5D%5E3%7D)
Ecell = 0.577 V
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Answer:
0.74 grams of methane
Explanation:
The balanced equation of the combustion reaction of methane with oxygen is:
it is clear that 1 mol of CH₄ reacts with 2 mol of O₂.
firstly, we need to calculate the number of moles of both
for CH₄:
number of moles = mass / molar mass = (3.00 g) / (16.00 g/mol) = 0.1875 mol.
for O₂:
number of moles = mass / molar mass = (9.00 g) / (32.00 g/mol) = 0.2812 mol.
- it is clear that O₂ is the limiting reactant and methane will leftover.
using cross multiplication
1 mol of CH₄ needs → 2 mol of O₂
??? mol of CH₄ needs → 0.2812 mol of O₂
∴ the number of mol of CH₄ needed = (0.2812 * 1) / 2 = 0.1406 mol
so 0.14 mol will react and the remaining CH₄
mol of CH₄ left over = 0.1875 -0.1406 = 0.0469 mol
now we convert moles into grams
mass of CH₄ left over = no. of mol of CH₄ left over * molar mass
= 0.0469 mol * 16 g/mol = 0.7504 g
So, the right choice is 0.74 grams of methane
I would say D. Let me know if i am wrong.
Answer:
B- fission breaks apart nuclei; fusion puts them back together
