Answer:
100 g
Explanation:
From the question given above, the following data were obtained:
Original amount (N₀) = 400 g
Time (t) = 4 years
Half-life (t½) = 2 years
Amount remaining (N) =?
Next, we shall determine the number of half-lives that has elapse. This can be obtained as follow:
Time (t) = 4 years
Half-life (t½) = 2 years
Number of half-lives (n) =?
n = t / t½
n = 4 / 2
n = 2
Thus, 2 half-lives has elapsed.
Finally, we shall determine the amount remaining of the radioactive isotope. This can be obtained as follow:
Original amount (N₀) = 400 g
Number of half-lives (n) = 2
Amount remaining (N) =?
N = 1/2ⁿ × N₀
N = 1/2² × 400
N = 1/4 × 400
N = 0.25 × 400
N = 100 g
Thus, the amount of the radioactive isotope remaing is the 100 g.
(3.5mol)(24.106 g/1mol c6h6) =84.371 g C6H6
Chem bonds by exchanged charges create new characteristics as a result
The lewis structure is helpful in showing how the bonding between atoms of a molecule are. The lewis structure of ammonia would be that the nitrogen atom will share three pairs of electron with the three hydrogen atoms leaving nitrogen to have 1 lone pair.<span />
Answer:
<u>a</u><u>.</u><u> </u><u>True</u><u>.</u>
Explanation:
Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.
weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C
take an example of <u>e</u><u>t</u><u>h</u><u>a</u><u>n</u><u>o</u><u>l</u><u>:</u>
<u>
</u>
<u>
</u>
<u>B</u><u>y</u><u> </u><u>o</u><u>z</u><u>o</u><u>n</u><u>o</u><u>l</u><u>y</u><u>s</u><u>i</u><u>s</u><u>:</u>
Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.
take an example of propanol:
if it undergoes ozonolysis, it gives ethanal and methanal.