In the laboratory a student combines 40.6 mL of a 0.113 M copper(II) sulfate solution with 26.4 mL of a 0.329 M copper(II) iodid
1 answer:
Answer : The final concentration of copper(II) ion is, 0.198 M
Explanation :
First we have to calculate the moles of and
.
Moles of = Moles of
= 0.00459 mol
and,
Moles of = Moles of
= 0.00869 mol
Now we have to calculate the total moles of copper(II) ion and total volume of solution.
Total moles copper(II) ion = 0.00459 mol + 0.00869 mol
Total moles copper(II) ion = 0.0133 mol
and,
Total volume of solution = 40.6 mL + 26.4 mL = 64.0 mL = 0.067 L (1 L = 1000 mL)
Now we have to calculate the final concentration of copper(II) ion.
Thus, the final concentration of copper(II) ion is, 0.198 M
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Answer:
NH3 —> base
We will see ammonia attach itself to silver and not the other way around, silver therefore accepts electrons. So, silver is a Lewis acid and ammonia is a Lewis base, so this reaction is complex as ammonia reacts with the silver cation .
Any group that contains an empty orbital in its valence shell can be considered a Lewis acid, but the bases do not differ in the Brønsted system from those in the Lewis system. Thus, acids are divided into types, including:
Simple Cations
Theoretically, they can be considered as acids and therefore we can also expect their acidic strength to increase for the following reasons:
1. By increasing the positive charge on the cation.
2. Increasing the charge of the nucleus from one atom to another in any period in the periodic table.
3. By decreasing the radius of the cation.
4. Decrease in the number of electron shells in the cation.
Therefore, the acidity of cations of any chain of transition elements increases with increasing atomic number when their positive charge is equal. There are some examples that show the reaction of cations as Lewis acids:
A) the reaction of the silver ion with ammonia
b) aluminum ion hydration
c) Adding alcohol to the lithium ion
D) formation of a ferric cyanide ion
I hope I helped you^_^
The structures of the isomers and the m/z values of their peaks are not given in the question. The complete question is provided in the attachment
Answer:
Compound 2 (2,5-dimethylhexane) will not have the peaks at 29 and 85 m/z
Explanation:
The fragmentation of molecules by electron ionization of mass spectrometer occurs according to Stevenson's Rule, which states that "The most probable fragmentation is the one that leaves the positive charge on the fragment with the lowest ionization energy". This is much like the Markovnikov's Rule in organic chemistry which has predicted the formation of most stable carbocation and the addition of hydrogen halide to it.
The mass spectra of compound 1 (2,4-dimethylhexane) will contain all the m/z values mentioned in the question. Each peak indicate towards homologous series of fragmentation product of the compound 1. The first peak can be attributed to ethyl carbocation (m/z = 29), with the increase of 14 units the next peak indicates towards propyl carbocation (m/z = 43) and onwards until molecular ion peak of 114 m/z.
Compound 2 (2,5-dimethylhexane) structure shows that the cleavage of C-C bond will not yield a stable ethyl and hexyl carbocation. Hence, no peaks will be observed at 29 and 85 m/z. The absence of these two peaks can be used to distinguish one isomer from the other.
