Answer:
468 h
Explanation:
Let's consider the reduction of chromium (III) to chromium that occurs in the electrolytic purification.
Cr³⁺ + 3 e⁻ → Cr
We can establish the following relations.
- 1 kg = 1,000 g
- The molar mass of Cr is 52.00 g/mol
- 1 mole of Cr is deposited when 3 moles of electrons circulate
- The charge of 1 mole of electrons is 96,468 c (Faraday's constant)
- 1 A = 1 c/s
- 1 h = 3,600 s
The hours that will take to plate 11.5 kg of chromium onto the cathode if the current passed through the cell is held constant at 38.0 A is:

Answer:
The answer is (c) binding themselves to invade antigens
1s2 2s2 2p6 2s2 3p6 4s2 3d5
Carbon has an electron configuration of 1s^2 2s^2 2p^2. During sp hybridization, one s and one p orbital of carbon combine to form two sp hybrid orbitals.