Answer:
a
Explanation:
the answer is a i'm pretty sure it might be wrong tho i'm sorry
Answer:
A rule of thumb is that 1.5 lbs. of baking soda per 10,000 gallons of water will raise alkalinity by about 10 ppm. If your pool's pH is tested below 7.2, add 3-4 pounds of baking soda. If you're new to adding pool chemicals, start by adding only one-half or three-fourths of the recommended amount.
B, Rocks are always being recycled into different forms.
Answer:
The values for spin quantum number +1/2 and - 1/2
Explanation:
Principal quantum number denoted by (n) is used to describe the shell or orbits that electrons are found. Principal quantum number can assume a value of n= 1,2, 3, 4,5............ which indicates K, L, M, N, O shell respectively.
To know the maximum number of electrons in each shell, the formula (2n²) can be used. The letter 'n' denotes the values of principal quantum number 1,2,3,4
For example
- n=1 (K shell) has maximum number of 2 electrons
- n=2 (L shell) has the maximum number of 8 electrons
- n=3 (M shell) has the maximum number of 18 electrons
- n=4 (N shell) has the maximum number of 32 electrons
All the electron in each shell will have a spin quantum number of +1/2 and - 1/2. One electron in each degenerate orbital will spin up (+1/2) while the other electron will spin down (-1/2).
Answer:
a) galvanic cell
b)electrolytic cell
c) i) K=6.27x10'34
ΔG°=198790 J
ii) K=3.58x10'-34
ΔG°= 191070 J
d) E°=0.278 v
ΔG°= -26827 J
Explanation:
a) There are two kinds of an electrochemical cell, the first is called "galvanic cells", and the second "electrolytic cell".
The fuel cells are capable of produce electric energy through chemical reactions. These reactions are often spontaneous. So, the galvanic cell has a negative value for Gibbs free energy.
b) The electrolytic cell increases the value of Gibbs energy, to positive values, due to the reactions are not spontaneous.
c) i) look image attached
ii) k = look image attached
ΔG° = -nFE° = - 6 X 95500 J/vmole x (-0.33 v)
ΔG° =-191070
d) E°= 0.0592 v/n x lg K
E°= 0.0592V / 1 X log 5.0X10'4
E°= 0.278 v
ΔG° = -nFE° = -1 x 96500 J/ vmole x 0.278v
ΔG° = -26827 J
