Answer:PH2=0.994atm, mass of zinc=0.606g
Explanation:
Equation for the reaction is written as;
Zn(s) + 2HCl(aq) - ZnCl2(aq) + H2(g)
V= 240ml = 0.240L
T= 30.8oC = 303.8K
Ptotal= 1.036atm
Pwater = 32mmHg at 30oC = 0.042atm
Therefore
PH2 = Ptotal - Pwater
= 1.036-0.042atm
=0.994atm
But PV= nRT
n = PV/RT
= 0.994x0.240/0.0821x303.8
= 0.24/24.94
=0.009324moles
For the grams of zinc
n=mass in grams/molar mass
Mass in grams = no of moles x molar mass
=0.009324x65 = 0.606g
Answer:
E° = 0.00 V
E = 0.079 V
Explanation:
We can identify both half-reactions occurring in a concentration cell.
Anode (oxidation): Al(s) → Al³⁺(1.0 × 10⁻⁵ M) + 3 e⁻ E°red = -1.66 V
Cathode (reduction): Al³⁺(0.100 M) + 3 e⁻ → Al(s) E°red = -1.66 V
The global reaction is:
Al(s) + Al³⁺(0.100 M) → Al³⁺(1.0 × 10⁻⁵ M) + Al(s)
The standard cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.
E° = E°red, cat - E°red, an = -1.66 V - (-1.66 V) = 0.00 V
To calculate the cell potential (E) we have to use the Nernst equation.
E = E° - (0.05916/n) .log Q
where,
n: moles of electrons transferred
Q: reaction quotient
E = 0.00 V - (0.05916/3) .log (1.0 × 10⁻⁵/0.100)
E = 0.079 V
Answer:
energy for the body to work
Explanation:
Answer:
four electrons
Explanation:
Let us attempt to write the electronic configuration of carbon in the ground state. This electronic configuration will now be;
C- 1s2 2s2 2p2
The outermost principal energy level of carbon is the n=2 level which houses the 2s2 and 2p2, making a total of four electrons in the outermost principal energy level of an atom of carbon in the ground state.
Answer:
1.5M
Explanation:
Molarity = moles/volume
0.60 mol / 0.40 L = 1.5 M
