Q1)
molarity is defined as the number of moles of solute in 1 L solution
the number of moles of LiNO₃ - 0.38 mol
volume of solution - 6.14 L
since molarity is number of moles in 1 L
number of moles in 6.14 L - 0.38 mol
therefore number of moles in 1 L - 0.38 mol / 6.14 L = 0.0619 mol/L
molarity of solution is 0.0619 M
Q2)
the mass of C₂H₆O in the solution is 72.8 g
molar mass of C₂H₆O is 46 g/mol
number of moles = mass present / molar mass of compound
the number of moles of C₂H₆O - 72.8 g / 46 g/mol
number of C₂H₆O moles - 1.58 mol
volume of solution - 2.34 L
number of moles in 2.34 L - 1.58 mol
therefore number of moles in 1 L - 1.58 mol / 2.34 L = 0.675 M
molarity of C₂H₆O is 0.675 M
Q3)
Mass of KI in solution - 12.87 x 10⁻³ g
molar mass - 166 g/mol
number of mole of KI = mass present / molar mass of KI
number of KI moles = 12.87 x 10⁻³ g / 166 g/mol = 0.0775 x 10⁻³ mol
volume of solution - 112.4 mL
number of moles of KI in 112.4 mL - 0.0775 x 10⁻³ mol
therefore number of moles in 1000 mL- 0.0775 x 10⁻³ mol / 112.4 mL x 1000 mL
molarity of KI - 6.90 x 10⁻⁴ M
Answer:
Li
Explanation:
The phenomenon of wave particle duality was well established by Louis deBroglie. The wavelength associated with matter waves was related to its mass and velocity as shown below;
λ= h/mv
Where;
λ= wavelength of matter waves
m= mass of the particle
v= velocity of the particle
This implies that if the velocities of all particles are the same, the wavelength of matter waves will now depend on the mass of the particle. Hence; the wavelength of a matter wave associated with a particle is inversely proportional to the magnitude of the particle's linear momentum. The longest wavelength will then be obtained from the smallest mass of matter. Hence lithium which has the smallest mass will exhibit the longest DeBroglie wavelength
Given mass of tungsten, W = 415 g
Molar mass of tungsten, W = 183.85 g/mol
Calculating moles of tungsten from mass and molar mass:
The molarity is the number of moles in 1 L of the solution.
The mass of NH₃ given - 2.35 g
Molar mass of NH₃ - 17 g/mol
The number of NH₃ moles in 2.35 g - 2.35 g / 17 g/mol = 0.138 mol
The number of moles in 0.05 L solution - 0.138 mol
Therefore number of moles in 1 L - 0.138 mol / 0.05 L x 1L = 2.76 mol
Therefore molarity of NH₃ - 2.76 M
