Answer:
The anode half reaction is : 
Explanation:
In electrochemical cell, oxidation occurs in anode and reduction occurs in cathode.
In oxidation, electrons are being released by a species. In reduction, electrons are being consumed by a species.
We can split the given cell reaction into two half-cell reaction such as-
Oxidation (anode):
Reduction (cathode): 
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overall: 
So the anode half reaction is :
Answer:
CO = zero
CO2 =1 bar
O2 = 2.02 bar
Explanation:
We are given
initial pressure of CO = 1bar
total pressure = 3.52 bar
so initial pressure of O2 = 3.52 - 1 = 2.52 bar
the reaction is
2CO + O2 → 2CO2
using the unitary method
2 moles of CO2 → 1 mole of O2
1 bar of CO → 
(required)
but we have more oxygen present , that means CO is the limiting reagent
- Final pressure of CO will be zero as it is the limiting reagent so it will be consumed completely
- 1 bar of CO →
of CO2 - 2.52 bar O2 (initially) - 0.5 bar (reacted) = 2.02bar O2
The moles of chromium (iii) nitrate produced is calculated as follows
write the equation for reaction
3 Pb(NO3)2 + 2 Cr = 2 Cr(NO3)3 + 3 Pb
by use of mole ratio between Pb(NO3)2 to Cr(NO3)3 which is 3 : 2 the moles of Cr(NO3)3 is therefore
= 0.85 x2 /3 = 0.57 moles
Answer:
Final pressure = 2.3225 atm
Amontons’s law states that
At constant volume and number of molecules, the pressure of a given mass of gas is directly proportional to its temperature
Explanation:
Temperature causes increased excitement of gas molecules increasing the number of collisions with the walls of the container which is sensed as increase in pressure
Amontons’s law: P/T = Constant at constant V and n
That is P1/T1 = P2/T2
Where temperature is given in Kelvin
Hence T1 of 10°C = 273.15 + 10 = 283.15K
Also temperature T2 of 40°C = 313.15 K
Hence
P2 = (P1/T1)×T2 = (2.1/283.15)×313.15 = 2.3225 atm
Usually concentrations are expressed as molarity, or moles of solute per liter solution. First, convert the mass of bromide ion to moles. The molar mass of bromine is 79.904 g/mol.
Moles of bromine = 65 mg * 1 g/1000 mg * 1 mol/79.904 g = 8.135×10⁻⁴ moles
Next, convert the mass of seawater to volume using the density.
Volume of seawater = 1 kg * 1 m³/ 1,025 kg * 1000 L/1 m³ = 0.976 L
Thus,
Molarity = 8.135×10⁻⁴ moles/0.976 L = 8.335×10⁻⁴ M
