The solution for this problem is:
The information given lacks but nevertheless the answer is:
So, the total heat free by dissolving the solute was 1386 + 32 = 1417 J
Then, dissolving of the solute will have released -1417 J. So, per gram of Al2
(SO4)3 dissolved:
-1417 J / 25 g = -56.7 J/g
Translating that to a per mole: -56.7 J/g X 342 g/mol = -1.94X10^4 J/mol = -19
kJ/mol = this would be Delta Hsoln
Answer:
d. 3 signals: a singlet, a doublet, and a septet
Explanation:
In this case, we can start with the structure of 
. When we draw the molecule we will obtain <u>2-methoxypropane</u> (see figure 1).
In 2-methoxypropane we will have three signals. The signal for the 
groups in the left, the 
and the in the right. Lets analyse each one:
-) in the right
In this carbon, we dont have any hydrogen as neighbors. Therfore we will have <u>singlet</u> signal in this carbon.
-)
In this case, we have 6 hydrogen neighbors ( the two methyl groups in the left). So, if we follow the <u>n + 1 rule</u> (where n is the amount of hydrogen neighbors):
For this carbon we will have a <u>septet</u>.
-) in the left
In this case we have only 1 hydrogen neighbor (the hydrogen in
). So, if we use the n+1 rule we will have:
We will have a doublet
With all this in mind the answer would be:
<u>d. 3 signals: a singlet, a doublet, and a septet
</u>
<u />
See figure 2 to further explanations
Answer:
Object 2 with a density of 0.68
Explanation:
object 2 has a lower density than water so it will float
Answer:
The compound you will use is the Dibasic phosphate
Explanation:
Simple phosphate buffer is used ubiquitously in biological experiments, as it can be adapted to a variety of pH levels, including isotonic. This wide range is due to phosphoric acid having 3 dissociation constants, (known in chemistry as a triprotic acid) allowing for formulation of buffers near each of the pH levels of 2.15, 6.86, or 12.32. Phosphate buffer is highly water soluble and has a high buffering capacity,
In this case the most efficient way is to disolve the dibasic compound which in the reaction with the water will form the monobasic phosphate.
To make the buffer you have to prepare the amount of distillate water needed, disolve the dibasic phospate, and then adjust with HCl or NaOH depending on the pH needed.
Answer:
CN^- is a strong field ligand
Explanation:
The complex, hexacyanoferrate II is an Fe^2+ specie. Fe^2+ is a d^6 specie. It may exist as high spin (paramagnetic) or low spin (diamagnetic) depending on the ligand. The energy of the d-orbitals become nondegenerate upon approach of a ligand. The extent of separation of the two orbitals and the energy between them is defined as the magnitude of crystal field splitting (∆o).
Ligands that cause a large crystal field splitting such as CN^- are called strong field ligands. They lead to the formation of diamagnetic species. Strong field ligands occur towards the end of the spectrochemical series of ligands.
Hence the complex, Fe(CN)6 4− is diamagnetic because the cyanide ion is a strong field ligand that causes the six d-electrons present to pair up in a low spin arrangement.
