This is an incomplete question, the table is attached below.
Answer : The correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.
Explanation :
As we know that the intermolecular force of attraction play an important role in the interaction of solute-solute, solute-solvent and solvent solvent solution.
In the solution A, the solute-solute and solvent-solvent interactions are weak. So, their solute-solvent interaction will be strong. That means, the solution will be more exothermic.
In the solution C, the solute-solute and solvent-solvent interactions are strong. So, their solute-solvent interaction will be weak. That means, the solution will be more endothermic.
Thus, the correct ranking of the solution from most exothermic to most endothermic will be: A, B and C.
The number of grams of Ag2SO4 that could be formed is 31.8 grams
<u><em> calculation</em></u>
Balanced equation is as below
2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)
- Find the moles of each reactant by use of mole= mass/molar mass formula
that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles
moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles
- use the mole ratio to determine the moles of Ag2SO4
that is;
- the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles
- The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles
- AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles
<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>
that is 0.102 moles x 311.8 g/mol= 31.8 grams
Answer:
The possible structures are ketone and aldehyde.
Explanation:
Number of double bonds of the given compound is calculated using the below formula.
=Number of double bonds
= Number of carbon atoms
= Number of hydrogen atoms
= Number of nitrogen atoms
The number of double bonds in the given formula - 
The number of double bonds in the compound is one.
Therefore, probable structures is as follows.
(In attachment)
The structures I and III are ruled out from the probable structures because the signal in 13C-NMR appears at greater than 160 ppm.
alkene compounds I and II shows signal less than 140 ppm.
Hence, the probable structures III and IV are given as follows.
The carbonyl of structure I appear at 202 and ketone group of IV appears at 208 in 13C, which are greater than 160.
Hence, the molecular formula of the compound having possible structure in which the signal appears at greater than 160 ppm are shown aw follows.
Answer : The correct option is, (a) paramagnetic with two unpaired electrons.
Explanation :
According to the molecular orbital theory, the general molecular orbital configuration will be,
![(\sigma_{1s}),(\sigma_{1s}^*),(\sigma_{2s}),(\sigma_{2s}^*),(\sigma_{2p_z}),[(\pi_{2p_x})=(\pi_{2p_y})],[(\pi_{2p_x}^*)=(\pi_{2p_y}^*)],(\sigma_{2p_z}^*)](https://tex.z-dn.net/?f=%28%5Csigma_%7B1s%7D%29%2C%28%5Csigma_%7B1s%7D%5E%2A%29%2C%28%5Csigma_%7B2s%7D%29%2C%28%5Csigma_%7B2s%7D%5E%2A%29%2C%28%5Csigma_%7B2p_z%7D%29%2C%5B%28%5Cpi_%7B2p_x%7D%29%3D%28%5Cpi_%7B2p_y%7D%29%5D%2C%5B%28%5Cpi_%7B2p_x%7D%5E%2A%29%3D%28%5Cpi_%7B2p_y%7D%5E%2A%29%5D%2C%28%5Csigma_%7B2p_z%7D%5E%2A%29)
As there are 14 electrons present in the given configuration.
The molecular orbital configuration of molecule will be,
![(\sigma_{1s})^2,(\sigma_{1s}^*)^2,(\sigma_{2s})^2,(\sigma_{2s}^*)^2,(\sigma_{2p_z})^2,[(\pi_{2p_x})^1=(\pi_{2p_y})^1],[(\pi_{2p_x}^*)^0=(\pi_{2p_y}^*)^0],(\sigma_{2p_z}^*)^0](https://tex.z-dn.net/?f=%28%5Csigma_%7B1s%7D%29%5E2%2C%28%5Csigma_%7B1s%7D%5E%2A%29%5E2%2C%28%5Csigma_%7B2s%7D%29%5E2%2C%28%5Csigma_%7B2s%7D%5E%2A%29%5E2%2C%28%5Csigma_%7B2p_z%7D%29%5E2%2C%5B%28%5Cpi_%7B2p_x%7D%29%5E1%3D%28%5Cpi_%7B2p_y%7D%29%5E1%5D%2C%5B%28%5Cpi_%7B2p_x%7D%5E%2A%29%5E0%3D%28%5Cpi_%7B2p_y%7D%5E%2A%29%5E0%5D%2C%28%5Csigma_%7B2p_z%7D%5E%2A%29%5E0)
The number of unpaired electron in the given configuration is, 2. So, this is paramagnetic. That means, more the number of unpaired electrons, more paramagnetic.
Hence, the correct option is, (a) paramagnetic with two unpaired electrons.
Answer: The substance is in the gas phase only in region 5
The substance is in both the liquid and the solid phase in region 2
The substance is in only the liquid phase in region 3
The melting point is the temperature at region 2
The boiling point is the temperature at region 4
Explanation:
