Answer:
B. electrons possess the largest charge-to-mass ratio among the subatomic particles listed in the four choices.
Explanation:
Consider the mass of each particle. Express the masses in atomic mass units:
- Protons: approximately 1.007 amu each;
- Neutrons: approximately 1.009 amu each;
- Electrons: approximately 0.0005 amu each.
Similarly, consider the charge on each particle. Express the charges in multiples of the fundamental charge:
- Protons: +1 e;
- Neutrons: 0;
- Electrons: -1 e.
Calculate the charge-to-mass ratio for the three species:
- Protons: approximately
; - Neutrons: 0;
- Electrons: approximately
.
Almost all nuclei contain protons and neutrons. The only exception is the hydrogen-1 nucleus, which contains only one proton and no neutron. The mass of the nucleus is approximately the same as the sum of its components' masses. The extra neutron will only add to the mass of the nucleus (the denominator) without contributing to the charge (the numerator.) As a result, the charge-to-mass ratio of nuclei will be positive but no greater than the charge-to-mass ratio of protons.
Among the particles in the four choices, the charge-to-mass ratio is the greatest for electrons.
Answer:
Halogens
Explanation:
From the given choices, the halogens will have the smallest radius within the same period.
The size of an atom is estimated by the atomic radius. This is taken as half of the inter-nuclear distance between two covalently bonded atoms of non-metallic elements or half of the distance between two nuclei in the solid state.
- Across a period in the periodic table, atomic radii decrease progressively from left to right.
- Down a group from top to bottom, atomic radii increase progressively due to the addition of successive shells.
Since halogen is the right most group from the choices given, it will have the smallest radius.
Electrolysis of water<span> is the </span><span>decomposition reaction, because from one molecule (water) two molecules (hydrogen and oxygen) are produced. Water is separeted into two molecules:
</span>Reaction of reduction at cathode: 2H⁺(aq) + 2e⁻<span> → H</span>₂(g<span>).
</span><span><span>Reaction of oxidation at anode: 2H</span></span>₂<span><span>O(l) → O</span></span>₂<span><span>(g) + 4H</span></span>⁺(<span><span>aq) + 4e</span></span>⁻.<span><span>
</span></span>
Answer:
The final pressure of the gas is 0.915atm
Explanation:
We have to apply the Charles Gay Lussac Law, where the pressure changes directly proportional to absolute T°
- No change in volume
- The same moles in both situations
P1 / T1 = P2 / T2
0.991 atm / 342K = P2 / 316k
(0.991 atm / 342K) . 316K = P2
0.915 atm = P2
They work by lowering the activation energy
