Answer: Ag: 2.79x10^22 atoms
C atoms: 6.02x10^23 C atoms
Explanation:
5.0 grams of Ag is 0.0464 moles of Ag. (0.0464 moles)*(6.02x10^23 atoms/mole) = 2.79x10^22 atoms Ag
(0.50 moles)*(6.02x10^23 molecules/mole)*(2 C atoms/molecule) = 6.02x10^23 atoms of C
Answer:
O=O bond
Explanation:-
Note it down that the bond having highest Hydrogen enthalpy has strongest bond.
Now
- O=O(495KJ/mol)
- O-O(146KJ/mol)
- O-H(467KJ/mol)
- H-H(432KJ/mol)
Hence

So in your question that ask to calculate the Ph result of the resulting solution if 26 ml of 0.260 M HCI(aq) is added to the following substance. The the result are the following:
A. The result is pH= 14-pOH
B. There are 10ml of 0.26m HCL excees in this reaction so the answer is log(H)+
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: I think the answer is 1
i just learned this about two weeks ago
Explanation: