Answer:
208.8018 g/Mol you conver the answer by weight and moles
B, D and E
Explanation:
conversion factors
1c = 8oz
1pt = 2c
1qt = 2pt
For A and B
ounces to cup = 160/8 = 20c
cup to pints = 20c / 2c = 10pt
pint to quarts = 10pt/2pt = 5qt
B applies as four 1-quart and two 1-pt = 5 1-quart
Considering C
4 8oz = 32oz
160-32 = 128oz /8 = 16c/2 = 8pints C does not apply
considering D
8 * 8 = 64oz
160 - 64 = 96oz/8 = 12c/2 = 6pt/2 = 3qt
So D applies
E applies
Answer:
Isotopes are atoms of the same element that have different numbers of neutrons but the same number of protons and electrons. The difference in the number of neutrons between the various isotopes of an element means that the various isotopes have different masses.
C. Aluminum (Al) oxidized, zinc (Zn) reduced
<h3>Further explanation</h3>
Given
Metals that undergo oxidation and reduction
Required
A galvanic cell
Solution
The condition for voltaic cells is that they can react spontaneously, indicated by a positive cell potential.
![\large {\boxed {\bold {E ^ ocell = E ^ ocatode -E ^ oanode}}}](https://tex.z-dn.net/?f=%5Clarge%20%7B%5Cboxed%20%7B%5Cbold%20%7BE%20%5E%20ocell%20%3D%20E%20%5E%20ocatode%20-E%20%5E%20oanode%7D%7D%7D)
or:
E ° cell = E ° reduction-E ° oxidation
For the reaction to occur spontaneously (so that it E cell is positive), the E° anode must be less than the E°cathode
If we look at the voltaic series:
<em>Li-K-Ba-Ca-Na-Mg-Al-Mn- (H2O) -Zn-Cr-Fe-Cd-Co-Ni-Sn-Pb- (H) -Cu-Hg-Ag-Pt-Au </em>
The standard potential value(E°) from left to right in the voltaic series will be greater, so that the metal undergoing an oxidation reaction (acting as an anode) must be located to the left of the reduced metal (as a cathode)
<em />
From the available answer choices, oxidized Al (anode) and reduced Zn (cathode) are voltaic/galvanic cells.
Answer:
by moving between defined energy levels