I will have to say that this statement is true
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
Read more about Temperature
brainly.com/question/13439286
1st one is Science and the 2nd one is Bias
Hey There!
At neutralisation moles of H⁺ from HCl = moles of OH⁻ from Ca(OH)2 so :
0.204 * 42.8 / 1000 => 0.0087312 moles
Moles of Ca(OH)2 :
2 HCl + Ca(OH)2 = CaCl2 + 2 H2O
0.0087312 / 2 => 0.0043656 moles ( since each Ca(OH)2 ives 2 OH⁻ ions )
Therefore:
Molar mass Ca(OH)2 = 74.1 g/mol
mass = moles of Ca(OH)2 * molar mass
mass = 0.0043656 * 74.1
mass = 0.32 g of Ca(OH)2
Hope that helps!