5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.
Calcium oxalate (CaC₂O₄) is obtained by the reaction of 5 g of potassium oxalate (K₂C₂O₄).
We can calculate the moles of CaC₂O₄ obtained considering the following relationships.
- The molar mass of K₂C₂O₄ is 184.24 g/mol.
- The mole ratio of K₂C₂O₄ to CaC₂O₄ is 2:1.
5 g of potassium oxalate react to produce 0.03 moles of calcium oxalate.
Learn more: brainly.com/question/15288923
Find volume of pillow
L=78cm
B=55cm
H=25cm
Now
Mass=5.5kg
Density of water=1000kg/m^3
As it is less than density of water it will float on water
<span>n this order, Ď=1.8gmL, cm=0.5, and mole fraction = 0.9
First, let's start with wt%, which is the symbol for weight percent. 98wt% means that for every 100g of solution, 98g represent sulphuric acid, H2SO4.
We know that 1dm3=1L, so H2SO4's molarity is
C=nV=18.0moles1.0L=18M
In order to determine sulphuric acid solution's density, we need to find its mass; H2SO4's molar mass is 98.0gmol, so
18.0moles1Lâ‹…98.0g1mole=1764g1L
Since we've determined that we have 1764g of H2SO4 in 1L, we'll use the wt% to determine the mass of the solution
98.0wt%=98g.H2SO4100.0g.solution=1764gmasssolution→
masssolution=1764gâ‹…100.0g98g=1800g
Therefore, 1L of 98wt% H2SO4 solution will have a density of
Ď=mV=1800g1.0â‹…103mL=1.8gmL
H2SO4's molality, which is defined as the number of moles of solute divided by the mass in kg of the solvent; assuming the solvent is water, this will turn out to be
cm=nH2SO4masssolvent=18moles(1800â’1764)â‹…10â’3kg=0.5m
Since mole fraction is defined as the number of moles of one substance divided by the total number of moles in the solution, and knowing the water's molar mass is 18gmol, we could determine that
100g.solutionâ‹…98g100gâ‹…1mole98g=1 mole H2SO4
100g.solutionâ‹…(100â’98)g100gâ‹…1mole18g=0.11 moles H2O
So, H2SO4's mole fraction is
molefractionH2SO4=11+0.11=0.9</span>
