Answer:
Particle Symbol Mass
electron e- 0.0005486 amu
proton p+ 1.007276 amu
neutron no 1.008665
Above question is incomplete. Complete question is attached below
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Solution:
Reduction potential of metal ions are provided below. Higher the value to reduction potential, greater is the tendency of metal to remain in reduced state.
In present case,
reduction potential of Au is maximum, hence it is least prone to undergo oxidation. Hence, it is
least reactive.
On other hand,
reduction potential of Na is minimum, hence it is most prone to undergo oxidation. Hence, it is
most reactive.
the bond in a molecule of Cl2 is a covalent bond
The new pressure is 81.675 torr
Since temperature and moles are held constant, we use Boyle's Law:
A gas law known as Boyle's law asserts that a gas's pressure is inversely proportional to its volume when it is held at a fixed temperature and of a given mass.
To put it another way, as long as the temperature and volume of the gas remain constant, the pressure and volume of the gas are inversely proportional to one another.
The Anglo-Irish chemist Robert Boyle proposed Boyle's law in the year 1662.
P1V1=P2V2. Simply plug in your values. The units can remain in torr. Converting to atmospheres is not needed.
(242 torr)(27.0 L)=P2(80.0 L)
P2=[(242)(27)]/80 = 81.675 torr
Hence The new pressure is 81.675 torr
Learn more about Boyle's Law here
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Answer: Li is the reducing agentg and O is the oxidizing agent.
Explanation:
1) The oxidizing agent is the one that is reduced and the reducing agent is the one that is oxidized.
2) The given reaction is:
4Li(s) + O₂ (g) → 2 Li₂O(s)
3) Determine the oxidation states of each atom:
Li(s): oxidation state = 0 (since it is alone)
O₂ (g): oxidation state = 0 (since it is alone)
Li in Li₂O (s) +1
O in Li₂O -2
That because 2× (+1) - 2 = 0.
4) Determine the changes:
Li went from 0 to + 1, therefore it got oxidized and it is the reducing agent.
O went from 0 to - 2, therefore it got reduced and it is the oxidizing agent.