The number of half -lives that has passed after 105 hours for krypton-79 that has half-life of 35 hours is calculated as below
if 1 half life = 35 hours
what about 105 hours = ? half-lives
= (1 half life x105 hours) /35 hours = 3 half-lives has passed after 105 hours
Less reactive than Group<span> I </span>elements<span>. The reasoning for this is because it is </span>more<span> difficult to lose two electrons compared to losing just </span>one<span> electron. They mostly React with water to form alkaline solutions. ...Now This is because the smaller an atom the closer the outer electrons are to the nucleus.</span>
This is a redox reaction, meaning reduction-oxidation reaction. This represents the reaction in one side of the electrode in an electrolysis set-up. First, we find the oxidation number of Cu in CuSO4:
(ox. # of Cu)+ ox.# of S + 4(ox.# of oxygen) = 0
(ox. # of Cu) + (6) + 4(-2) = 0
ox. # of Cu = 2+
CuSO4 ---> Cu + SO42-
Cu2+ + SO42- ----> Cu + SO42-
Cu2+ -----> Cu + 2e- (net ionic reaction)
The stoichiometric equation would be 2 electrons per mole Copper. Copper has a molar mass of <span>63.5 g/mol. Then, it would only need 2 electrons.
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Answer:
It is an ether so the nomenclature rule for naming the above compound will be:Alkoxy alkane , generally Alkoxy represent the smaller alkyl group and as in the above question both the alkyl group are same so it's IUPAC Name will be: Ethoxy ethane or Diethylether both are correct
Explanation:
happy to help ya!
What we are give: Concentration of base (CB) = 3.4 ×
Then convert all volume in ml to L.
Volume of base (VB) 25.0ml = 0.025L
Volume of acid (VA) 16.6ml = 0.0166L
Now that we have everything we use the formula CAVA=CBVB.
Make 'CA' the subject then solve.
CA=
