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).
c. A full s subshell is able to shield a newly filled p subshell from the nucleus, making the first electron in a p subshell easy to remove.
Explanation:
From the given options, a full s-sublevel is able to shield a newly filled p-subshell from the nucleus thereby making the first electron in a p-subshell easy to remove is correct.
What is ionization energy?
Ionization energy is a measure of the readiness of an atom to lose an electron.
First ionization energy is the energy required to remove the most loosely held electron in the gas phase.
The size of an atom/element depends on the number of electrons it contains. The more the electrons, the larger its size.
- The larger an atom becomes the lesser the ionization energy needed to remove the first electron from its outermost shell.
Electron - electron repulsion occurs when two electrons in the same sub-level repels one another.
Shielding effect is the ability of the inner electrons to protect the outer electrons from the pull of the nuclear charge.
In option C, a s-subshell has a greater shielding effect than the p,d and f sub-shell in that order.
A newly introduced electron in the p-sublevel will be loosely held and easier to remove.
Learn more:
First ionization energy brainly.com/question/2153804
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We have M1V1=M2V2. Here M1 is to be found, V1=25ml=(25/1000)L &
M2=1.0M, V2=18ml=(18/1000)L.
Therefore M1=(M2V2)V1
so after putting values we get answer 0.72M.
Hope it helps.
Explanation:
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