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:
If they are pushing off the wall, it would be B, as they are going faster. If they are slowing down, it would probably be A, gradually getting slower.
Explanation:
Answer: Freezing point of a solution will be 
Explanation:
Depression in freezing point is given by:
= Depression in freezing point
i= vant hoff factor = 1 (for non electrolyte)
= freezing point constant = 
m= molality
Weight of solvent (benzene)= 1480 g =1.48 kg
Molar mass of solute (octane) = 114.0 g/mol
Mass of solute (octane) = 220 g
Thus the freezing point of a solution will be
The rate constant : k = 9.2 x 10⁻³ s⁻¹
The half life : t1/2 = 75.3 s
<h3>Further explanation</h3>
Given
Reaction 45% complete in 65 s
Required
The rate constant and the half life
Solution
For first order ln[A]=−kt+ln[A]o
45% complete, 55% remains
A = 0.55
Ao = 1
Input the value :
ln A = -kt + ln Ao
ln 0.55 = -k.65 + ln 1
-0.598=-k.65
k = 9.2 x 10⁻³ s⁻¹
The half life :
t1/2 = (ln 2) / k
t1/2 = 0.693 : 9.2 x 10⁻³
t1/2 = 75.3 s
