Answer:
The correct option is: D.act as a Lewis base in water.
Explanation:
Ammonia is a hydride of nitrogen with the chemical formula NH₃. It is a colorless gas with a characteristic pungent smell. According to the VSEPR theory, it has a trigonal pyramidal structure.
In water, ammonia acts as a Lewis base due to the presence of lone pair on the nitrogen atom. Lewis bases are electron pair or lone pair donors.
b. magnetic centers
c. domains
d. poles
Answer:
The correct answer is c. domains
Explanation:
In terms of Chemistry, there is a concept called the electron domain and it has to do with the number of lone pairs or bond locations that are located around a particular atom in a molecule. The electron domains are also called the electron groups.
The answer is Lipids. Hope it helps.
b. activation energy.
c. catalyst energy.
d. chemical energy.
Activation energy is the energy needed to start a chemical reaction. The correct answer to the given question is option b. activation energy.
The energy needed to get a reaction started is called the activation energy. It is the minimum amount of energy required for a chemical reaction to occur. Activation energy is needed to break the existing bonds in reactant molecules and form new bonds to produce product molecules.
The correct answer to the given question is option b. activation energy.
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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.