Answer:
I can't say that it is definitely write.
HHH
H-C-C-C
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 dk how ur school works, but you'll most likely have to take that science again to get the credits u need to graduate
Explanation:
Answer:
0.1113 mol
Explanation:
Data Given:
no. of atoms of CH₄= 6.70 x 10²² atoms
no. of moles of methane (CH₄) = ?
Solution:
we will find no. of moles of methane (CH₄)
Formula used
no. of moles = no. of atoms / Avogadro's number
Where
Avogadro's number = 6.022 x 10²³
Put values in above equation
no. of moles = 6.70 x 10²² atoms / 6.022 x 10²³ (atoms/mol)
no. of moles = 0.1113 mol
So,
There are 0.1113 moles of methane.
