D. The final substance in Beaker A is a mixture and in Beaker B is a pure substance.
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
Answer:
Condensation methods from colloidal particles by aggregation of molecules or ions. Examples of colloids are really in common in evryday life, eg. Mayonnaise, butter, milk, gelatin, paper etc..
Every colloid consists of two parts :colloidal particles and the dispersing medium.
<span>An element is made of only one kind of atom, while a compound contains the atoms of two or more elements. For example, gold is made of only one kind of atom, so it is an element, but water is a mix of hydrogen and oxygen, so it is a compound.</span>
Answer:
1. 0.125 mole
2. 42.5 g
3. 0.61 mole
Explanation:
1. Determination of the number of mole of NaOH.
Mass of NaOH = 5 g
Molar mass of NaOH = 23 + 16 + 1
= 40 g/mol
Mole of NaOH =?
Mole = mass /molar mass
Mole of NaOH = 5/40
Mole NaOH = 0.125 mole
2. Determination of the mass of NH₃.
Mole of NH₃ = 2.5 moles
Molar mass of NH₃ = 14 + (3×1)
= 14 + 3
= 17 g/mol
Mass of NH₃ =?
Mass = mole × molar mass
Mass of NH₃ = 2.5 × 17
Mass of NH₃ = 42.5 g
3. Determination of the number of mole of Ca(NO₃)₂.
Mass of Ca(NO₃)₂ = 100 g
Molar mass of Ca(NO₃)₂ = 40 + 2[14 + (3×16)]
= 40 + 2[14 + 48]
= 40 + 2[62]
= 40 + 124
= 164 g/mol
Mole of Ca(NO₃)₂ =?
Mole = mass /molar mass
Mole of Ca(NO₃)₂ = 100 / 164
Mole of Ca(NO₃)₂ = 0.61 mole
