Answer:
Explanation:
We are to carefully sketch a curve that relates to the potential energy of two O atoms versus the distance between their nuclei.
From the diagram, O2 have higher potential energy than the N2 molecule. Because on the periodic table, the atomic size increases from left to right on across the period, thus O2 posses a larger atomic size than N2 atom.
Therefore, the bond length formation between the two O atoms will be larger compared to that of the two N atoms.
Answer:
110ml
Explanation:
<em>Using the dilution equation, C1V1 = C2V2</em>
<em>Where C1 is the initial concentration of solution</em>
<em>C2 is final concentration of solution</em>
<em>V1 is intital volume of solution</em>
<em>V2 is final volume of solution.</em>
From the question , C1=6M, C2=0.5M, V1=10ml, V2=?



volume of water added = final volume -initial volume
= 120-10
=110ml
<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:
19,26 kJ
Explanation:
The work done when a gas expand with a constant atmospheric pressure is:
W = PΔV
Where P is pressure and ΔV is the change in volume of gas.
Assuming the initial volume is 0, the reaction of 500g of Zn with H⁺ (Zn(s) + 2H⁺(aq) → Zn²⁺(aq) + H₂(g)) produce:
500,0g Zn(s)×
×
= 7,648 moles of H₂
At 1,00atm and 303,15K (30°C), the volume of these moles of gas is:
V = nRT/P
V = 7,648mol×0,082atmL/molK×303,15K / 1,00atm
V = 190,1L
That means that ΔV is:
190,1L - 0L = <em>190,1L</em>
And the work done is:
W = 1atm×190,1L = 190,1atmL.
In joules:
190,1 atmL×
= <em>19,26 kJ</em>
I hope it helps!