Your answer would be,
Molarity = moles of solute/volume of solution we needed, 29.22(g)(mol) of NaCI
= 29.22(g)/58.44(g)(mol^-1)(1)/1(L)
= 0.500(mol)(L^-1)
Hope that helps!!!
<h3>Answer:</h3>
There is One electrophilic center in acetyl chloride.
<h3>Explanation:</h3>
Electrophile is defined as any specie which is electron deficient and is in need of electrons to complete its electron density or octet. The main two types of electrophiles are those species which either contain positive charge (i.e. NO₂⁺, Cl⁺, Br⁺ e.t.c) or partial positive charge like that contained by the sp² hybridized carbon of acetyl chloride shown below in attached picture.
In acetyl chloride the partial positive charge on sp² hybridized carbon is generated due to its direct bonding to highly electronegative elements *with partial negative charge) like oxygen and chlorine, which tend to pull the electron density from carbon atom making it electron deficient and a good electrophile for incoming nucleophile as a center of attack.
Answer:
region 2 and region 3
Explanation:
you can tell by the color of the land my friends^^
Answer: Elements in Group 2
Explanation: The periodic table was arranged by Dmitri Mendeleev specifically around similarites in their chemical behaviors. He found that as atomic number increases, at some point an element starts to react in a manner similar to a previous one. When that happened, he would place the larger element under the smaller one, and eventually noticed a periodicity in the table. Elements in a column (Groups) had similiar chemical properties. We know today that these similarities are due to the electron configuration, and that these configurations repeat themselves. He left gaps in the table when he could find an existing element with properties similar to others in that group. I big leap of faith, but it worked. Elements for those missing boxes were eventually discovered.
Answer:
The answer is 3-Phenylpropanoic acid (see attached structure)
Explanation:
From spectral data:
3005 cm-1 ⇒ carboxylic acid (broad band)
1670 cm-1 ⇒ C=C
1603 cm-1 ⇒ Aromatic C-C bond
H NMR frequency at 2.6 ppm, singlet, ⇒ OH with no surrounding protons, possible deshielding (clearer investigation of spectrum would be expedient).
Hence, our C9H10O2 compound has an aromatic ring and carboxylic acid group attached to it.