Answer:
1.99 x 10²³ formula units
Explanation:
Given parameters:
Mass of AgF = 42.15g
Unknown:
The amount of formula units
Solution:
To solve this problem, we set out by find the number of moles in this compound from the given mass.
Number of moles = 
molar mass of AgF = 107.9 + 19 = 126.9g/mol
Number of moles =
= 0.33 moles
1 mole of a substance = 6.02 x 10²³ formula units
0.33 moles of AgF = 0.33 x 6.02 x 10²³ = 1.99 x 10²³ formula units
Answer:
Molar mass of vitamin K = 450.56\frac{g}{mol}[/tex]
Explanation:
The freezing point of camphor = 178.4 ⁰C
the Kf of camphor = 37.7°C/m
where : m = molality
the relation between freezing point depression and molality is
Depression in freezing point = Kf X molality
Where
Kf = cryoscopic constant of camphor
molality = moles of solute dissolved per kg of solvent.
putting values
2.69°C = 37.7°C/m X molality
molality = 0.0714 mol /kg

moles of vitamin K = 0.0714X0.025 = 0.00178 mol
we know that moles are related to mass and molar mass of a substance as:

For vitamin K the mass is given = 0.802 grams
therefore molar mass = 
Explanation:
Principle Quantum Numbers : It describes the size of the orbital and the energy level. It is represented by n. Where, n = 1,2,3,4....
Azimuthal Quantum Number : It describes the shape of the orbital. It is represented as 'l'. The value of l ranges from 0 to (n-1). For l = 0,1,2,3... the orbitals are s, p, d, f...
s = 1 orbital
p = 3 orbitals
d = 5 orbitals
f = 7 orbitals
For n = 4
l = 0 to (n-1) = 0 to 3 = (4s , 4p , 4d , 4f)
Number of subshells = 4
Number of orbitals = 1 + 3 + 5 + 7 = 16
The maximum number of electrons the n = 4 shell can contain:
Each orbital can holds upto two electrons, then 16 orbitals will have :

32 is the maximum number of electrons the n = 4 shell can contain
Answer:
30 mL VOLUME OF 3.0 M HCl SHOULD BE USED BY THE STUDENT TO MAKE A 1.80 M IN 50 mL OF HCl.
Explanation:
M1 = 3.00 M
M2 = 1.80 M
V2 = 50 .0 mL = 50 /1000 L = 0.05 L
V1 = unknown
In solving this question, we know that number of moles of a solution is equal to the molar concentration multiplied by the volume. To compare two samples, we equate both number of moles and substitute for the required component.
So we use the equation:
M1 V1 = M2 V2
V1 = M2 V2 / M1
V2 = 1.80 * 0.05 / 3.0
V2 = 0.09 /3.0
V2 = 0.03 L or 30 mL
To prepare the sample of 1.80 M HCl in 50.0 mL from a 3.0 M HCl, 30 mL volume should be used.