Answer:
Case 1:
X = Any element from Group I
i) H
ii) Li
iii) Na
iv) K
v) Rb
vi) Cs
Y = 1
Case 2:
X = Any element from Group II
i) Be
ii) Mg
iii) Ca
iv) Sr
v) Ba
vi) Ra
Y = 2
Case 3:
X = Any element from Group III
i) B
ii) Al
iii) Ga
iv) In
v) Ti
Y = 3
Explanation:
The general formula given is as follow,
XCly
So, if X has +1 oxidation state, then it will require only one Cl atom with oxidation number -1 to form a neutral compound, therefore, y = 1.
If X has +2 oxidation state, then it will require two Cl atoms with oxidation number -1 to form a neutral compound, therefore, y = 2.
If X has +3 oxidation state, then it will require three Cl atoms with oxidation number -1 to form a neutral compound, therefore, y = 3.
Answer: A. -396 kJ
Explanation:
The standard enthalpy of formation or standard heat of formation of a compound is the change of enthalpy during the formation of 1 mole of the substance from its constituent elements, with all substances in their standard states.
Reversing the reaction, changes the sign of 
On multiplying the reaction by 2, enthalpy gets multiplied by 2:
Thus the enthalpy change for the reaction is -396 kJ.
Answer:
26 g left undisolved which gives you the amount that would crystallize
Explanation:
Step 1: Data given
A solid mixture consists of 44.2g of KNO3 (potassium nitrate) and 7.8g of K2SO4 (potassium sulfate).
The mixture is added to 130. g of water
KNO3 has a solubility of 14 g solute/ 100 g water
Step 2: Calculate mass of KNO3 in 130 grams of water
X g of KNO3/ 130 g water= 14 g of KNO3/ 100 g water
X = 14*1300/100 = 18.2
X= 18.2 g (this is the amount of solute that dissolves at 0 degrees C in 130 g water)
Step 3: Calculate amount KNO3 that will crystallize
Of the 44.2g of KNO3 in 130 g of water at 0 degrees celsiusonly 18.2 can dissolve, this means:
44.2g - 18.2g = 26 g left undisolved which gives you the amount that would crystallize
