the answer will be b)
Metals tend to lose electrons.
Atoms have the natural tendency to fill their outermost shells.
All of the above.
Answer: Metals tend to gain electrons.
Explanation: Hope this helps!!! Have an amazing day!
Answer:
Non-metals tend to lose electrons
impure substance.
mixture.
pure substance.
Answer:
D. pure substance.
Explanation:
Answer:
:D
Explanation:
Answer:
oxidation reaction
Explanation:
Rusting is the oxidation of iron in the presence of air and moisture. Rusting in a chemical change in which iron combines with oxygen to form iron oxide. The molar mass of iron oxide is greater than iron. So the mass of an object after rusting is greater than its initial mass.
Answer:the standard reduction potential (E°red) for the reduction half-reaction of Pb^4+(aq) to Pb^2+(aq) is 1.50 V.
Explanation:The given chemical reaction is:
Pb^4+(aq) + 2 Ce^3+(aq) -> Pb^2+(aq) + 2 Ce^4+(aq)
The standard cell potential (E°cell) for this electrochemical reaction is 0.06 V.
The standard cell potential for a galvanic cell can be calculated using the Nernst equation:
E°cell = E°cathode - E°anode
In this reaction, Pb^4+(aq) is being reduced to Pb^2+(aq), so it is the reduction half-reaction, and Ce^3+(aq) is being oxidized to Ce^4+(aq), so it is the oxidation half-reaction.
The standard reduction potentials (E°) for the half-reactions are as follows:
For the reduction half-reaction:
Pb^4+(aq) + 2 e^- -> Pb^2+(aq) E°red = x (we'll solve for x)
For the oxidation half-reaction:
2 Ce^3+(aq) -> 2 Ce^4+(aq) + 2 e^- E°red = 1.44 V (This value is usually given)
Now, plug these values into the Nernst equation:
E°cell = E°cathode - E°anode
0.06 V = x - 1.44 V
Now, solve for x:
x = 0.06 V + 1.44 V
x = 1.50 V
So, the standard reduction potential (E°red) for the reduction half-reaction of Pb^4+(aq) to Pb^2+(aq) is 1.50 V.
(2) 2.0 L of 2.0 M HCl(aq)
(3) 3.0 L of 0.50 M HCl(aq)
(4) 4.0 L of 0.50 M HCl(aq)
Answer:
2
Explanation: