Answer:
Most common oxidation state of the chalcogens is -2, most common oxidation state of the halogens is -1.
Explanation:
For atomic radii, the chalcogens have a larger atomic radii than the halogens
This is because atomic radii decreases across the period due to increase in nuclear charge.
For ionic radii the chalcogens also have larger ionic radii than the halogens. This is because the chalcogens always carry a -2 charge compared to halogens that carry a -1 charge. Since -2 is the most common oxidation state for chalcogens and -1 is the most common oxidation state for the halogens.
In terms of oxidation states, the halogens show a higher value of common oxidation state -1 while for chalcogens is -2 even though +2, +4 and +6 oxidation states are also well known.
First ionization energy of halogens is greater than that of the chalcogens due to greater effective nuclear charge.
The second ionization energy of chalcogens is greater than that of the halogens.
The net ionic equation of the reaction could be determined by cancelling out the like ions between both sides of the reaction. These ions are called spectator ions. They are called as such because they do not actively participate in the reaction. The spectator ions are Na+ and Cl-. When you cancel those, the equation would become letter D.
Answer:
Explanation:
This question appear incomplete because of the absence of options. However, places where silver recovery would happen in the United States are states where huge mining of copper, lead, zinc or gold are been mined. Examples of such places include Arizona, California and Oregon; where copper (and some other metals in some cases like gold in Oregon) is heavily mined and silver is recovered as a byproduct.
Answer:
[NaOH} = 0.4 M
Explanation:
In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.
(H₂SO₄, is considered strong, but the first deprotonation is weak)
2NaOH + H₂SO₄ → Na₂SO₄ + 2H₂O
As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.
In the equivalence point we know mmoles of base = mmoles of acid
Let's finish the excersise with the formula
25 mL . M NaOH = 28.2 mL . 0.355M
M NaOH = (28.2 mL . 0.355M) / 25 mL → 0.400