Their dark matter detector witnessed the rarest event ever recorded: the radioactive decay of xenon-124. ... The supreme fine-tuning and clean measurements allowed by XENON1T enabled scientists to witness xenon-124 decay away at a rate that far exceeds the life of the universe.
Hope that helps.
Answer:
Approximately
.
Explanation:
The gallium here is likely to be produced from a
solution using electrolysis. However, the problem did not provide a chemical equation for that process. How many electrons will it take to produce one mole of gallium?
Note the Roman Numeral "
" next to
. This numeral indicates that the oxidation state of the gallium in this solution is equal to
. In other words, each gallium atom is three electrons short from being neutral. It would take three electrons to reduce one of these atoms to its neutral, metallic state in the form of
.
As a result, it would take three moles of electrons to deposit one mole of gallium atoms from this gallium
solution.
How many electrons are supplied? Start by finding the charge on all the electrons in the unit coulomb. Make sure all values are in their standard units.
.
.
Calculate the number of electrons in moles using the Faraday's constant. This constant gives the size of the charge (in coulombs) on each mole of electrons.
.
It takes three moles of electrons to deposit one mole of gallium atoms
. As a result,
of electrons would deposit
of gallium atoms
.
Answer:
theory is diffrent from law
Explanation:
a Theory can never be proven to be true nd a law can usually be expressed
Answer:
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)
Explanation:
Step 1: Data given
Volume = 500 mL = 0.500 L
The concentration sodium sulfate = 2.104 M
Step 2: The equation
Na2SO4 → 2Na+ + SO4^2-
For 1 mol Na2SO4 we have 2 moles sodium ion (Na+) and 1 mol sulfate ion (SO4^2-)
Step 3: Calculate the concentration of the ions
[Na+] = 2*2.104 M = 4.208 M
[SO4^2-] = 1*2.104 M = 2.104 M
The concentration of species in 500 mL of a 2.104 M solution of sodium sulfate is 4.208 M sodium ion and 2.104 M sulfate ion. (option E)