The molar mass of a, b and c at STP is calculated as below
At STP T is always= 273 Kelvin and ,P= 1.0 atm
by use of ideal gas equation that is PV =nRT
n(number of moles) = mass/molar mass therefore replace n in the ideal gas equation
that is Pv = (mass/molar mass)RT
multiply both side by molar mass and then divide by Pv to make molar mass the subject of the formula
that is molar mass = (mass x RT)/ PV
density is always = mass/volume
therefore by replacing mass/volume in the equation by density the equation
molar mass=( density xRT)/P where R = 0.082 L.atm/mol.K
the molar mass for a
= (1.25 g/l x0.082 L.atm/mol.k x273k)/1.0atm = 28g/mol
the molar mass of b
=(2.86g/l x0.082L.atm/mol.k x273 k) /1.0 atm = 64 g/mol
the molar mass of c
=0.714g/l x0.082 L.atm/mol.K x273 K) 1.0atm= 16 g/mol
therefore the
gas a is nitrogen N2 since 14 x2= 28 g/mol
gas b =SO2 since 32 +(16x2)= 64g/mol
gas c = methaneCH4 since 12+(1x4) = 16 g/mol
scandium is a transition metal and shows a valency of 3.
Answer:
1.12M
Explanation:
Given parameters:
Volume of solution = 2.5L
Mass of Calcium phosphate = 600g
Unknown:
Concentration = ?
Solution:
Concentration is the number of moles of solute in a particular solution.
Now, we find the number of moles of the calcium phosphate from the given mass;
Formula of calcium phosphate = Ca₃PO₄
molar mass = 3(40) + 31 + 4(16) = 215g/mol
Number of moles of Ca₃PO₄ = 
= 2.79moles
Now;
Concentration = 
Concentration = 
= 1.12M
Answer:
2Na=Ca(OH)000.1 AgBr=2KF 2KBr=LiNO
Answer:
I think the answer is the 3rd one
