Answer : The cell potential for this reaction is 0.50 V
Explanation :
The given cell reactions is:
The half-cell reactions are:
Oxidation half reaction (anode): 
Reduction half reaction (cathode): 
First we have to calculate the cell potential for this reaction.
Using Nernest equation :
![E_{cell}=E^o_{cell}-\frac{2.303RT}{nF}\log \frac{[Zn^{2+}]}{[Pb^{2+}]}](https://tex.z-dn.net/?f=E_%7Bcell%7D%3DE%5Eo_%7Bcell%7D-%5Cfrac%7B2.303RT%7D%7BnF%7D%5Clog%20%5Cfrac%7B%5BZn%5E%7B2%2B%7D%5D%7D%7B%5BPb%5E%7B2%2B%7D%5D%7D)
where,
F = Faraday constant = 96500 C
R = gas constant = 8.314 J/mol.K
T = room temperature = 
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = +0.63 V
= cell potential for the reaction = ?
![[Zn^{2+}]](https://tex.z-dn.net/?f=%5BZn%5E%7B2%2B%7D%5D)
= 3.5 M
![[Pb^{2+}]](https://tex.z-dn.net/?f=%5BPb%5E%7B2%2B%7D%5D)
= 
Now put all the given values in the above equation, we get:
Therefore, the cell potential for this reaction is 0.50 V
Answer:
c) parallel to the wire
Explanation:
The concept is based on right-hand thumb rule that states that the thumb points in the direction of the current and the fingers will represent the direction of the lines of magnetic force.
As the current is moving in the east-west direction. Using the right-hand thumb rule, the direction of the magnetic field at a point below the wire is from north to south. So, when viewed from the west end, the magnetic field at a point will be anti- clockwise.
Hence, the correct answer is "b) anti- clockwise".
Answer:
48.8%
Explanation:
The reaction has a 1:1 mole ratio so;
Number of moles of benzoic acid reacted = mass/molar mass = 3.8 g/122.12 g/mol = 0.03 moles
So;
0.03 moles of methyl benzoate is formed in the reaction
Mass of methyl benzoate formed = 0.03 moles * 136.15 g/mol = 4.1 g
percent yield = actual yield/theoretical yield * 100/1
percent yield = 2.0 g/4.1 g * 100 = 48.8%
They all don’t, they also can have positive charges like LiOH (Lithium Hydroxide)
The enthalpy of combustion of 1 mole of benzene is 3169 kJ/mol .
The first step in answering this question is to obtain the balanced thermochemical equation of the reaction. The thermochemical equation shows the amount of heat lost or gained.
The thermochemical equation for the combustion of benzene is;
2 C6H6(l) + 15 O2(g) → 12 CO2(g) + 6 H2O(g) ΔrH° = -3169 kJ/mol
We can see that 1 mole of benzene releases about 3169 kJ/mol of heat.
Learn more: brainly.com/question/13164491
