1.95 or 2 is the molarity of a 45.3g sample of KNO3 (101g) dissolved in enough water to make a 0.225L solution.
The correct answer is option b
Explanation:
Data given:
mass of KN
= 45.3 grams
volume = 0.225 litre
molarity =?
atomic mass of KNO3 = 101 grams/mole
molarity is calculated by using the formula:
molarity = 
first the number of moles present in the given mass is calculated as:
number of moles = 
number of moles = 
0.44 moles of KNO3
Putting the values in the equation of molarity:
molarity = 
molarity = 1.95
It can be taken as 2.
The molarity of the potassium nitrate solution is 2.
Answer:
Explanation:
Ksp(BaSO4)=1.07×10−10
BaSO₄ → Ba²⁺ + SO₄²⁻
1.07×10⁻¹⁰ = ( Ba²⁺) × ( SO₄²⁻)
but Ba²⁺ = 1.3×10⁻² M
1.07×10⁻¹⁰ = 1.3×10⁻² M × ( SO₄²⁻)
( SO₄²⁻) = 1.07×10⁻¹⁰ / 1.3×10⁻² = 0.823 × 10⁻⁸ M
while Ksp(CaSO4)=7.10×10−5
CaSO₄ → Ca²⁺ + SO₄²⁻
7.10×10⁻⁵ = 2.0×10⁻² × ( SO₄²⁻)
( SO₄²⁻) = 7.10×10⁻⁵ / 2.0×10⁻² = 3.55 × 10⁻³ M
comparing the concentration of sulfate ions, Ba²⁺ cation will precipitate first because the Ba²⁺ requires 0.823 × 10⁻⁸ M sodium sulfate which less compared the about needed by CaSO₄
Answer: Option (C) is the correct answer.
Explanation:
Chemical formula of a secondary amide is R'-CONH-R, where R and R' can be same of different alkyl or aryl groups. Here, the hydrogen atom of amide is attached to more electronegative oxygen atom of the C=O group.
Therefore, the hydrogen atom will be more strongly held by the electronegative oxygen atom. As a result, there will be strongly hydrogen bonded in the liquid phase of secondary amide.
Whereas chemical formula of nitriles is RCN, ester is RCOOR' and acid chlorides are RCOCl. As no hydrogen bonding occurs in any of these compounds because hydrogen atom is not being attached to an electronegative atom.
Thus, we can conclude that secondary amides are strongly hydrogen bonded in the liquid phase.
