Answer:
The correct answer is option (B) Two molecules of H2 and a molecule of O2 must collide at one time
Explanation:
2H2(g) + O2(g) → 2 H2O(g)
2 1 2
Considering the given reaction, for water molecule to be form, two molecules of hydrogen and one molecule of oxygen must collide at the same time. The product obtained from the collision is two molecules of water as steam. 2 H2O(g)
Answer:
sp³
Explanation:
Number of hybrid orbitals = ( V + S - C + A ) / 2
Where
H is the number of hybrid orbitals
V is the valence electrons of the central atom = 5
S is the number of single valency atoms = 4
C is the number of cations = 1
A is the number of anions = 0
For PCl₄⁺
Applying the values, we get:
H = ( 5+4-1+0) / 2
= 4
<u>This corresponds to sp³ hybridization.</u>
Answer:
youre gonna have to include the answers for me to help
Explanation:
Answer:
Part A. The half-cell B is the cathode and the half-cell A is the anode
Part B. 0.017V
Explanation:
Part A
The electrons must go from the anode to the cathode. At the anode oxidation takes place, and at the cathode a reduction, so the flow of electrons must go from the less concentrated solution to the most one (at oxidation the concentration intends to increase, and at the reduction, the concentration intends to decrease).
So, the half-cell B is the cathode and the half-cell A is the anode.
Part B
By the Nersnt equation:
E°cell = E° - (0.0592/n)*log[anode]/[cathode]
Where n is the number of electrons being changed in the reaction, in this case, n = 2 (Sn goes from S⁺²). Because the half-reactions are the same, the reduction potential of the anode is equal to the cathode, and E° = 0 V.
E°cell = 0 - (0.0592/2)*log(0.23/0.87)
E°cell = 0.017V
Answer:
yes answer os Na because it's electronic configuration is 1s^2,2s^2,2p^6,3s^1
