Answer:
5w2
Explanation:
Electrons are distributed around the nucleus and occupy almost all the volume of the atom.
The nucleus is made of electrons and protons.
The nucleus is made of protons, electrons, and neutrons.
The answer is (B) Electrons are distributed around the nucleus and occupy almost all the volume of the atom.
Answer:
The answer to your question is 2 moles of Carbon
Explanation:
Carbon-12 means it has a molecular mass of 12g / mol. Mol is defined as the number of grams of an atom in one mol.
To solve this problem, we use proportions
12 g of Carbon ---------------- 1 mol of Carbon
24 g of Carbon -------------- x
x = ( 24 g x 1 mol ) / 12 g
x = 24 / 12
x = 2 mol of Carbon
Answer:
2 moles
Explanation:
just took the test :D
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inventing stories?
Plz Help ;_; :v
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Answer:
One word for this is, raconteur - it means a person skilled in telling anecdotes. anecdotist. narrator, or storyteller
Explanation:
Answer:
ions and polar molecules
The potassium nuclei have been unstable and results in the spontaneous decay without any external force. Thus, option 3 is correct.
Potassium has been the metal of group 1 of the periodic table. It has been consisted of 19 protons and 20 neutrons in the nucleus.
The nuclei decay has been achieved when there has been loss of the protons in the nuclei, resulting in the radioactive isotopes.
The potassium nuclei have been unstable and results in the spontaneous decay without any external force. Thus, option 3 is correct.
Learn more about potassium nuclei, here:
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.