Answer:
a. -0.63 V
b. No
Explanation:
Step 1: Given data
- Standard reduction potential of the anode (E°red): -1.33 V
- Minimum standard cell potential (E°cell): 0.70 V
Step 2: Calculate the required standard reduction potential of the cathode
The galvanic cell must provide at least 0.70V of electrical power, that is:
E°cell > 0.70 V [1]
We can calculate the standard reduction potential of the cathode (E°cat) using the following expression.
E°cell = E°cat - E°an [2]
If we combine [1] and [2], we get,
E°cat - E°an > 0.70 V
E°cat > 0.70 V + E°an
E°cat > 0.70 V + (-1.33 V)
E°cat > -0.63 V
The minimum E°cat is -0.63 V and there is no maximum E°cat.
(NH4)3PO4 :
N = 14 * 3 = 42
H = 1 * 12 = 12
P = 31 * 1 = 31
O = 16 * 4 = 64
-------------------------
42+12+31+64 = 149 g / mol
Hope this helps!.
Answer:
160000000 ml is 42267.528 gallons of water
Answer:
d. is the hydrostatic pressure produced on the surface of a semi-permeable membrane by osmosis.
Explanation:
Osmosis -
It is the flow of the molecules of solvent from a region of higher concentration towards the region of lower concentration via a semipermeable membrane , is known as osmosis.
Osmotic pressure -
It refers to the minimum amount of pressure , which is required to be applied to the solution in order to avoid the flow of pure solvent via the semipermeable membrane , is referred to as osmotic pressure.
Or in simple terms ,
Osmotic pressure is the pressure applied to resists the process of osmosis.
Hence ,
From the given options in the question,
The correct option regarding osmotic pressure is d.
