Answer:
Two non bonded electron pairs and four bonded electron pairs
Explanation:
An image of the compound as obtained from chemlibretext is attached to this answer.
The ion ICl4- ion, is an AX4E2 ion. This implies that there are four bond pairs and two lone pairs of electrons. As expected, the shape of the ion is square planar since the lone pairs are found above and below the plane of the square. This is clear from the image attached.
The basis of finding the answer to this problem is to know the electronic configuration of Fluorine. That would be: <span>[He] 2s</span>²<span> 2p</span>⁵. The valence electrons, which are the outermost electrons of the atom, are the ones that participate in bonding. <em>Since the highest orbital for F is 2p, that means the highest energy occupied would be 2.</em>
Answer:
Amino >Methoxy > Acetamido
Explanation:
Bromination is of aromatic ring is an electrophilic substitution reaction. The attached functional group to the benzene ring activates or deactivate the aromatic ring towards electrophilic substitution reaction.
The functional group which donates electron to the benzene ring through inductive effect or resonance effect activates the ring towards electrophilic substitution reaction.
The functional group which withdraws electron to the benzene ring through inductive effect or resonance effect deactivates the ring towards electrophilic substitution reaction.
Among given, methoxy and amino are electron donating group. Amino group are stronger electron donating group than methoxy group. Acetamido group because of presence of carbonyl group becomes electron withdrawing group.
Therefore, decreasing order will be as follows:
Amino >Methoxy > Acetamido
CaCO3(s) ⟶ CaO(s)+CO2(s)
<span>
moles CaCO3: 1.31 g/100 g/mole CaCO3= 0.0131 </span>
<span>
From stoichiometry, 1 mole of CO2 is formed per 1 mole CaCO3,
therefore 0.0131 moles CO2 should also be formed.
0.0131 moles CO2 x 44 g/mole CO2 = 0.576 g CO2 </span>
Therefore:<span>
<span>% Yield: 0.53/.576 x100= 92 percent yield</span></span>
Answer:
Explanation:
In 150 ml of .06 g / ml solution , gram of iodine = 150 x .06 g = 9 g
Let volume of given concentration of .12 g / ml required be V
In volume V , gram of iodine = V x .12 g
According to question
V x .12 = 9 g
V = 9 / .12 = 75 ml
So, 75 ml of .12 g/ml will be taken and it is diluted to the volume of 150 ml to get the solution of required concentration .
