Answer:
The answer to your question is None of your answers is correct, maybe the data are wrong.
Explanation:
Data
Concentration 1 = C1 = 1 M
Volume 2 = 5 ml
Concentration 2 = 0.05 M
Volume 1 = x
To solve this problem use the dilution formula
Concentration 1 x Volume 1 = Concentration 2 x Volume 2
Solve for Volume 1
Volume 1 = (Concentration 2 x Volume 2)/ Concentration 1
Substitution
Volume 1 = (0.05 x 5) / 1
Simplification
Volume 1 = 0.25/1
Result
Volume 1 = 0.25 ml
Answer:
The classification is mentioned below for the particular topic.
Explanation:
- Whether we position 2 different beakers in such a single beaker through one clean edge of zinc-containing H₃Po₄ and another one with unflushed zinc.
- The zinc that was washed set to release hydrogen gas way quicker, unlike unventilated zinc.
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- Since fresh zinc complicates the cycle since, as a comparison to polluted zinc, there was little contact with either the reaction.
First determine the formal oxidation numbers:
N changes from +2 to +5 going from NO to (NO3)- O remains -2 the whole time Cr changes from +6 to +3
Now write the half reactions, balance the oxygens with the required number of waters and then balance the hydrogens with the required number of protons:
Oxidation half reaction:
NO(aq) + 2 H2O(l) ---> (NO3)-(aq) + 4 H+(aq) + 3 e-
Reduction half reaction:
(Cr2O7)2-(aq) + 14 H+(aq) + 6 e- ---> 2 Cr3+(aq) + 7 H2O(l)
Now balance the number of electrons on both sides and add them together:
2 NO(aq) + 4 H2O(l) ---> 2 (NO3)-(aq) + 8 H+(aq) + 6 e- (Cr2O7)2-(aq) + 14 H+(aq) + 6 e- ---> 2 Cr3+(aq) + 7 H2O(l) --------------------------------------... 2 NO(aq) + (Cr2O7)2-(aq) + 6 H+(aq) ---> 2 (NO3)-(aq) + 2 Cr3+(aq) + 3 H2O(l)
Notice that the charge is the same in both sides, which is an indication that the redox equation has been balanced correctly:
-2 + 6 = -2 + 2(+3) +4 = +4
According to Dalton; All matter is made up of small particles called atoms.
No, the two isotopes of lithium-6 and lithium-7 are not equally common.
The more plentiful isotope would be lithium-7.
This can be easily demonstrated by assuming that both isotopes were equally common. If that were the case, the average atomic mass would be (6 + 7)/2 = 6.5 amu. Now compare that value if they were both equal to the actual value found in nature. The value found in nature is 6.941 amu which is heavier than the 6.5 amu that would happen if they were equally common. Since the natural value is heavier, that means that there has to be more of the heavier isotope than there is of the lighter one. Therefore lithium-7 is more common than lithium-6.