Answer:they have the same number of electrons in the valence shell
Explanation:
Q1. Chemical, Physical, Physical, Physical
(l am not 100% sure about the 4th answer)
Q2. All of the above
For example we are going to use this unbalanced chemical reaction:
H₂ + O₂ → H₂O.
First, calculate number of atoms (hydrogens and oxygens) on left and right. There is two oxygen and two hydrogen on left and two hydrogen and one oxygen on right.
You can not change molecular formula of compound, only you can put coefficient in fron of compound to balance reaction.
Put 2 in front water to balance oxygen (now you have two oxygens on left and right). But now you have four hydrogens on right, so you must put 2 in fron hydrogen on the left.
2H₂ + O₂ → 2H₂O.
The molecular weight of water is <span>18.01528 g/mol.
So in 2.92 grams there are 2.92/</span>18.01528 = 0.1621 mol of particles.
1 mol contains 6,02214 × 10^<span>23 particles by definition.
So the nr of H2O molecules is </span>0.1621 * 6,02214 × 10^23 = 0,9761 × 10^23.
Every molecule has 2 H atoms, so you have to double that.
2* 0,9761 × 10^23 = 1.952 × 10^23.
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
