The basic units for density is
and any get of units that has those units in the proper place can be considered a density unit. The ones that has those specifically are A, B, E and F
Scientists use the physical and chemical properties to help them identify and classify matter. These physical and chemical properties are in a macro-perspective, in which these matter contains compounds, elements and atoms. Hence, matter can be classified in various ways, <span><span>
1. </span>Atomic number either atomic mass each element has</span>
<span><span>2. </span>By substance of that matter either pure substance or mixed substance</span> <span>
3. If they cannot reduce a certain substance into a much smaller quantified atomic structure then they they’ll use (2) to identify and classify it.</span>
<span>283.89 g/mol is the molar mass of tetraphosphorus decoxide</span>
Answer:
Total partial pressure, Pt = 821 mm Hg
Partial pressure of Helium, P1 = 105 mm Hg
Partial pressure of Nitrogen, P2 = 312 mm Hg
Partial pressure of Oxygen, P3 = ? mm Hg
According to Dalton's law of Partial pressures,
Pt = P1 + P2 + P3
So, <u>P3 = 404 mm Hg</u>
Answer:
a=28600J; b=90.6 J/K; c=402 torr
Explanation:
(a) considering the data given
Vapour pressure P1 =0 at Temperature T1 = 42.43˚C,
Vapour pressure P2 = 273.15 at Temperature T2= 315.58 K)
Using the Clausius-Clapeyron Equation
ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
In 760/140 = ΔH/8.314 J/mol/K × (1/315.58K -- 1/273.15K)
ΔH vap= +28.6 kJ/mol or 28600J
(b) using the Equation ΔG°=ΔH° - TΔS to solve forΔS.
Since ΔG at boiling point is zero,
ΔS =(ΔH°vap/Τb)
ΔS = 28600 J/315.58 K
= 90.6 J/K
(c) using ln (P2/P1) = (ΔH/R)(1/T2 - 1/T1)
ln P298 K/1 atm = 28600 J/8.314 J/mol/K × (1/298.15K - 1/315.58K)
P298 K = 0.529 atm
= 402 torr
