Answer: In an experiment, mice were fed glucose (C6H12O6) containing a small amount of radioactive carbon. The mice were closely monitored, and in a few minutes, radioactive carbon atoms showed up in _Carbon dioxide/Carbon IV oxide (CO2)_
Explanation: The end product of Glucose at the end of metabolism is Carbon dioxide, water and ATP.
C6H12O6 + 6O2 ----------> 6CO2 + 6H2O + ATP (Energy)
The carbon is converted to Carbon dioxide. So, the radioactive Carbon appears in the Carbon dioxide/Carbon IV oxide (CO2).
I’m not sure what this says could you translate it to English please?
When we have this equation:
CO(g) + Cl2(g) ↔ COCl2(g)
intial 0.147 0.175 0
change -X -X +X
final (0.147-X) (0.175-X) X
so from the ICE table, we substitute in Kc formula :(when we have Kc = 255)
Kc = [COCl2]/[CO][Cl2]
255= X / (0.147-X)(0.175-X)
255 = X / (X^2 - 0.322 X + 0.025725)
X = 0.13
∴[CO] = 0.147 - X = 0.147 - 0.13
= 0.017 m
The vapour pressure of a substance actually differs
relative to the temperature. So to know which compound has the vapour pressure
of 58 kilopascals at 65 degrees Celsius, we refer to standard tables of
temperature vs Vapour pressure.
From the tables or graph, the answer here is:
<span>2. ethanol</span>