The given solution of Mn²⁺ is 0.60 mg/mL.
Hence mass of Mn²⁺ in 5 mL of solution = 0.60 mg/mL x 5 mL = 3 mg
Molar mass of Mn = 54.9 g/mol
Hence, moles of Mn²⁺ = 3 x 10⁻³ g / 54.9 g/mol = 5.46 x 10⁻⁵ mol
The balanced equation for the reaction is,
2Mn²⁺ + 5KIO₄ + 3H₂O → 2MnO₄⁻ + 5KIO₃ + 6H⁺
The stoichiometric ratio between Mn²⁺ and KIO₄ is 2 : 5
Hence, moles of KIO₄ reacted = 5.46 x 10⁻⁵ mol x (5 / 2)
= 13.65 x 10⁻⁵ mol
Molar mass of KIO₄ = 230 g/mol
Hence needed mass of KIO₄ = 13.65 x 10⁻⁵ mol x 230 g/mol
= 0.031395 g
= 31.395 mg
≈ 31.4 mg
Answer:
The gas occupy 2406.4 mL at 80 K.
Explanation:
Given data:
Initial volume of gas = 752 mL
Initial temperature = 25 K
Final temperature = 80 K
Final volume = ?
Solution:
The given problem is solved by using charle's law.
V₁/T₁ = V₂/T₂
V₂ = V₁. T₂ /T₁
V₂ = 752 mL × 80 k / 25 K
V₂ = 60160 mL. k/25 K
V₂ = 2406.4 mL
The correct answer is that last one
Answer:
1. 0.073L
2. 0.028L
3. 0.014L
Explanation:
The volume for the different solutions are obtained as shown below:
1. Mole = 0.53mol
Molarity = 7.25M
Volume =?
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 0.53/7.25
Volume = 0.073L
2. 0.035mol from a 1.25M
Mole = 0. 035mol
Molarity = 1.25M
Volume =?
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 0.035/1.25
Volume = 0.028L
3. Mole = 0.0013mol
Molarity = 0.090M
Volume =?
Molarity = mole /Volume
Volume = mole /Molarity
Volume = 0.0013/0.090
Volume = 0.014L
