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Answer:
0.3477 grams
The conversion factor is 1/1000
Explanation:
347.7 mg = 0.3477 grams
1 mg = 1x10⁻³ g
1 g / 1000 mg . 347.7 mg = 0.3477 grams
Answer:
5.6L
Explanation:
At STP, the pressure and temperature of an ideal gas is
P = 1 atm
T = 273.15k
Volume =?
Mass = 9.5g
From ideal gas equation,
PV = nRT
P = pressure
V = volume
n = number of moles
R = ideal gas constant =0.082J/mol.K
T = temperature of the ideal gas
Number of moles = mass / molar mass
Molar mass of F2 = 37.99g/mol
Number of moles = mass / molar mass
Number of moles = 9.5 / 37.99
Number of moles = 0.25moles
PV = nRT
V = nRT/ P
V = (0.25 × 0.082 × 273.15) / 1
V = 5.599L = 5.6L
The volume of the gas is 5.6L
Answer:
The 
expression for the weak base equilibrium is:
![K_b=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
Explanation:
The expression of the equilibrium constant of base 
can be given as:
![K_c=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N][H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%5BH_2O%5D%7D)
]![K_b=K_c\times [H_2O]=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3DK_c%5Ctimes%20%5BH_2O%5D%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
As we know, water is pure solvent, we can put ![[H_2O]=1](https://tex.z-dn.net/?f=%5BH_2O%5D%3D1)
![K_b=K_c\times 1=\frac{[(CH_3)_3NH^+][OH^-]}{[(CH_3)_3N]}](https://tex.z-dn.net/?f=K_b%3DK_c%5Ctimes%201%3D%5Cfrac%7B%5B%28CH_3%29_3NH%5E%2B%5D%5BOH%5E-%5D%7D%7B%5B%28CH_3%29_3N%5D%7D)
So, the the expression for the weak base equilibrium is:
