<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>
Answer:
Original volume=3L
Explanation:
Assuming the gas in the ballon is ideal and the temperature does not vary with height we have the gas law as
PV=nRT
Here RHS is constant in both the conditions which means
PV=constant
Given P1=1 atm V1=V P2=0.75atm V2=4L
P1V1=P2V2
V=3L
The initial volume of the gas is 3L
Answer:
Graphite
Graphite is a non-metal and it is the only non-metal that can conduct electricity.
Let's rewrite the reaction for clarity:
2 SO₂(g) + O₂(g) ⇆ 2 SO₃(g) δhºrxn = –198 kj/mol
The equilibrium constant of a reaction is the ratio of the concentration its products to its reactants which are raised to their respective stoichiometric coefficients. For this reaction, the K would be
K = [SO₃]²/[SO₂]²[O₂]
To get a larger K, the products must be greater than the reactants. This means that the forward reaction must be favored to yield more of the product SO₃. There are different ways to do this: by manipulating the pressure, concentration or temperature.
For the concentration, you should add more amounts of the reactants. For the pressure, we should increase it. This is because the product side has only 2 moles of gas compared to 3 moles of gas in the reactants. So, it wall have more room for the product even at a higher pressure. Lastly, since the reaction is exothermic manifested by the negative sign of δhºrxn , the reaction would favor the forward reaction at high temperatures.
