Answer:
The product is significantly impure
Explanation:
In order to test for the purity of a specific sample that was synthesized, the melting point of a compound is measured. Basically speaking, the melting point identifies how pure a compound is. There are several cases that are worth noting:
- if the measured melting point is significantly lower than theoretical, e. g., lower by 3 or more degrees, we conclude that our compound contains a substantial amount of impurities;
- wide range in the melting point indicates impurities, unless it agrees with the theoretical range.
Since our compound is even 10 degrees Celsius lower than expected, it indicates that the compound is significantly impure.
Answer: They would both be pushed backward. (READ EXPLANATION)
Explanation:
Since your question has an error in it, I will provide two answers. If they were standing on their feet, they would not move. if they're both standing on skateboards, they would both move backwards because they have no friction on the ground. if both of them are standing on the same skateboard and pushing against each other, they would not move.
Isotopes are substances that have the same number of protons but differ in the number of neutrons. Hence, the pair of isotopes above should be of the same element. In the given choices, 14C is not an isotope of 14N. 206Pb is an isotope of 208 Pb. O2 and O3 differ in molecular formula but still made up of same kind of atom, hence they are allotropes, while 32S and 32S2- are not isotopes.
Density is defined as mass per unit volume
therefore density = mass ÷ volume
volume = mass÷density
volume in ml = 17 ÷ 3.291
Answer:
- <em><u>2I ⁻ → I₂ + 2e⁻</u></em> describes the <em><u>oxidation.</u></em>
- <u><em>Cl₂ + 2e⁻ → 2Cl ⁻</em></u> describes the <u><em>reduction</em></u>.
Explanation:
<em>Oxidation-reduction reaction</em> is the simulaneous oxidation and reduction of the substances and is represented by two half-reactions.
The <em>oxidation</em> half-reaction is the loss of electrons, with the consequent increase in the oxidation state by the oxidized substance.
In this case, the process that shows the loss of electrons is:
That reaction shows:
- Two I⁻ ions lose two electrons (one each) to be oxidized to I₂.
- The change in the oxidation number is from -1 to 0.
- Hence this half-reaction is the oxidation reaction.
On the other hand, the <em>reduction</em> half-reaction is the gain of electrons, with the consequent reduction of the oxidation state by the reduced substance.
In this case, the process that shows the gain of electrons is:
That reaction shows:
- Two Cl atoms gain two electrons (one each) to be reduced to Cl⁻.
- The change in the oxidation number is from 0 to - 1.
- Hence, this half-reaction is the reduction reaction.
<u>Summary:</u>
- <em>2I ⁻ → I₂</em> + 2e⁻ describes the oxidation.
- <em>Cl₂ + 2e⁻ → 2Cl ⁻</em> describes the reduction.
