Answer:
(c) 
Explanation:
The solubility product of a solid is the amount of solid dissociates into its respective ions in the solution. Thus more the value of the Ksp, the more is the salt soluble in the solvent.
So, Given that:-




The salt having highest value of Ksp is AgCN. So, it is most soluble.
Is this a question or a statement?
The balanced chemical reaction for the complete combustion of C4H10 is shown below:
C4H10 + (3/2)O2 --> 4CO2 + 5H2O
The enthalpy of formation are listed below:
C4H10: -2876.9 kJ/mol
O2: none (because it is pure substance)
CO2: -393.5 kJ/mol
H2O: -285.8 kJ/mol
The enthalpy of combustion is computed by subtracting the total enthalpy formation of the reactants from that of the products.
ΔHc = (4)(-393.5 kJ/mol) + (5)(-285.8 kJ/mol) - (-2876.9 kJ/mol)
= -<em>126.1 kJ</em>
Thus, the enthalpy of combustion of the carbon is -126.1 kJ.
<u>Answer:</u> The pressure that must be applied to the apparatus is 0.239 atm
<u>Explanation:</u>
To calculate the osmotic pressure, we use the equation for osmotic pressure, which is:

or,

where,
= osmotic pressure of the solution
i = Van't hoff factor = 1 (for non-electrolytes)
= mass of sucrose = 3.40 g
= molar mass of sucrose = 342.3 g/mol
= Volume of solution = 1 L
R = Gas constant = 
T = temperature of the solution = ![20^oC=[20+273]K=293K](https://tex.z-dn.net/?f=20%5EoC%3D%5B20%2B273%5DK%3D293K)
Putting values in above equation, we get:

Hence, the pressure that must be applied to the apparatus is 0.239 atm