Answer: a) Anode: 
Cathode: 
b) Anode : Cr
Cathode : Au
c) 
d) 
Explanation: -
a) The element Cr with negative reduction potential will lose electrons undergo oxidation and thus act as anode.The element Au with positive reduction potential will gain electrons undergo reduction and thus acts as cathode.
At cathode:
At anode:
b) At cathode which is a positive terminal, reduction occurs which is gain of electrons.
At anode which is a negative terminal, oxidation occurs which is loss of electrons.
Gold acts as cathode ad Chromium acts as anode.
c) Overall balanced equation:
At cathode: (1)
At anode: (2)
Adding (1) and (2)
d)
= -0.74 V
= 1.40 V
Using Nernst equation :
![E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Au^{3+}]}{[Cr^{3+}]^}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B0.0592%7D%7Bn%7D%5Clog%20%5Cfrac%7B%5BAu%5E%7B3%2B%7D%5D%7D%7B%5BCr%5E%7B3%2B%7D%5D%5E%7D)
where,
n = number of electrons in oxidation-reduction reaction = 3
= standard electrode potential = 2.14 V
![E_{cell}=2.14-\frac{0.0592}{3}\log \frac{[1.0}{[1.0]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3D2.14-%5Cfrac%7B0.0592%7D%7B3%7D%5Clog%20%5Cfrac%7B%5B1.0%7D%7B%5B1.0%5D%7D)
Thus the standard potential for an electrochemical cell with the cell reaction is 2.14 V.
Answer:
Molar mass of solute: 300g/mol
Explanation:
<em>Vapor pressure of pure benzene: 0.930 atm</em>
<em>Assuming you dissolve 10.0 g of the non-volatile solute in 78.11g of benzene and vapour pressure of solution was found to be 0.900atm</em>
<em />
It is possible to answer this question based on Raoult's law that states vapor pressure of an ideal solution is equal to mole fraction of the solvent multiplied to pressure of pure solvent:
Moles in 78.11g of benzene are:
78.11g benzene × (1mol / 78.11g) = <em>1 mol benzene</em>
Now, mole fraction replacing in Raoult's law is:
0.900atm / 0.930atm = <em>0.9677 = moles solvent / total moles</em>.
As mole of solvent is 1:
0.9677× total moles = 1 mole benzene.
Total moles:
1.033 total moles. Moles of solute are:
1.033 moles - 1.000 moles = <em>0.0333 moles</em>.
As molar mass is the mass of a substance in 1 mole. Molar mass of the solute is:
10.0g / 0.033moles = <em>300g/mol</em>
The mass of Ca(CN)2 : 92,11 g/mol
<h3>Further explanation</h3>
Given
0.0321 moles of Ca(CN)2
Required
The mass
Solution
The mole is the number of particles contained in a substance
1 mol = 6.02.10²³
Moles can also be determined from the amount of substance mass and its molar mass
mol(n) = mass(m) : MW(molecular weight)
Input the value :
mass = mol x MW Ca (CN)2
mass = 0.0321 x 92,11 g/mol
mass = 2.957 g
1) V(CH₄) = 0,376 L.T(CH₄) = 304 K.p(CH₄) = 1,5 atm 101325 Pa/atm = 151987,5 Pa = 151,9875 kPa.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.n(CH₄) = p · V ÷ R · T.n(CH₄) = 151,9875 kPa · 0,376 L ÷ 8,314 J/K· mol · 304 K.n(CH₄) = 0,0226 mol.V(CH₄) = n(CH₄) · Vm.V(CH₄) = 0,0226 mol · 22,4 dm³/mol.V(CH₄) = 0,506 dm³ = 0,506 L.
2) V(SO₂) = 5,2 L.p(SO₂) = 45,2 atm = 45,2 atm · 101,325 kPa/atm = 4579,89 kPa.T(SO₂) = 293 K.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.n(SO₂) = p · V ÷ R · T.n(SO₂) = 4579,89 kPa · 5,2 L ÷ 8,314 J/K· mol · 293 K.n(CH₄) = 9,77 mol.There is not enogh SO₂, 225 mol - 9,77 mol = 215,23 mol is needed.
3) p(He) = 3,50 atm · 101,325 kPa/atm = 354,63 kPa.V(He) = 4,00 L.n(He) = 0,410 mol.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.T = p · V ÷ R · n.T(He) = 354,63 kPa · 4,00 L ÷ 8,314 J/K· mol · 0,410 mol.T(He) = 416,14 K.n - amount of substance.
4) p(Ar) = 1,00 atm · 101,325 kPa/atm = 101,325 kPa.V(Ar) = 3,4 L.T(Ar) = 263 K.R = 8,314 J/K·mol.Use ideal gas law: p·V = n·R·T.n(Ar) = p · V ÷ R · T.n(Ar) = 101,325 kPa · 3,4 L ÷ 8,314 J/K· mol · 263 K.n(Ar) = 0,157 mol.n(Ar) = 0,157 mol + 2,5 mol = 2,657 mol.p(Ar) = 2,657 mol · 8,314 J/K· mol · 263 K ÷ 3,4 L.p(Ar) = 1708,74 kPa.
