1) Zn(CH₃COO)₂(s) + 2KOH(aq) = Zn(OH)₂(s) + 2CH₃COOK(aq)
Ksp{Zn(OH)₂}=1.2*10⁻¹⁷
2) Zn(CH₃COO)₂(s) + 2NaCN(aq) = Zn(CN)₂(s) + 2CH₃COONa(aq)
Ksp{Zn(CN)₂}=2.6*10⁻¹³
Ksp{Zn(OH)₂}<Ksp{Zn(CN)₂}
Zn(OH)₂ precipitates first
Answer: I & III
Explanation: Solutes are the substances which are minimum in quantity and which is required to dissolve in the solvent (which is larger in quantity) in order to make a solution.
In the asked question, it is given that the water is the solvent and from the given solutes we have to pick which would make an aqueous solution with the highest concentration of solute possible.
Thus the most appropriate answers could be the Ammonia and hexanol which can make the highest possible concentration of solute as ammonia is the gas which is highly soluble in water and hexanol is an alcohol which has an affinity for water. Thus the correct option is I & III
Answer:
Explanation:
From the information given:
Mass of carbon tetrachloride = 5 kg
Pressure = 1 bar
The given density for carbon tetrachloride = 1590 kg/m³
The specific heat of carbon tetrachloride = 0.84 kJ/kg K
From the composition, the initial volume of carbon tetrachloride will be:
= 0.0031 m³
Suppose 
is independent of temperature while pressure is constant;
Then:
The change in volume can be expressed as:
However; the workdone = -PdV
W = - 7.6 J
The heat energy Q = Δ h
Q = 84 kJ
The internal energy is calculated by using the 1st law of thermodynamics; which can be expressed as;
ΔU = ΔQ + W
ΔU = 84 kJ + ( -7.6 × 10⁻³ kJ)
ΔU = 83.992 kJ
Number of moles in the K2SO4 sample
= (16/1000)*1.04= 0.01664 mol
Number of moles in the Ba(NO3)2 sample
= (14.3/1000*0.880)= 0.01258 mol
Since the reaction is a 1:1 ratio between the two reactants, the limiting reagent is the one containing a smaller number of moles, namely Ba(NO3)2.
The molecular mass of BaSO4 is 137.3+(32.06+4*16.00)=233.4
Therefore the theoretical yield of Barium Sulphate is
233.4*0.01258=2.937 g
Actual yield = 2.60 g (given)
Therefore the percentage yield = 2.60/2.937=88.54%
Answer:
1. the limiting reagent is Barium Nitrate (Ba(NO3)2)
2. the theoretical yield is 2.94 g
3. the percentage yield is 88.5%
I apologize for the mistake previous to this update.
