Answer:
212.8 dm^3 or L
Explanation:
1 mole of any sub=6.02×10^23 molecules
X mole of O2=5.7×10^24 molecules
X mole=5.7×10^24/6.02×10^23
=9.5 mole
1 mole of any gas at stp=22.4 dm^3
Therefore, 9.5 mole of O2 will be 22.4×9.5
=212.8 dm^3
Answer:
(i)The mole fractions are :
(ii)
(iii)ΔG = 1.974kJ
Explanation:
The given equation is :
⇄
Let
be the number of moles dissociated per mole of 
Thus ,
<em>The initial number of moles of :</em>
+
⇄
+ 
And finally the number of moles of ![C[tex] is 0.9Thus ,[tex]3\alpha=0.9\\\alpha=0.3[tex]The final number of moles of:[tex]A = 1-2\alpha=1-2*0.3=0.4mol[tex] [tex]B=2(1-\alpha)=2(1-0.3)=1.4mol[tex][tex]D=1+2\alpha=1+2*0.3=1.6mol[tex]Thus , total number of moles are : 0.4+1.4+0.9+1.6=4.3(i)The mole fractions are : [tex]A=\frac{0.4}{4.3} \\=0.0930](https://tex.z-dn.net/?f=C%5Btex%5D%20is%200.9%3C%2Fp%3E%3Cp%3EThus%20%2C%3C%2Fp%3E%3Cp%3E%5Btex%5D3%5Calpha%3D0.9%5C%5C%5Calpha%3D0.3%5Btex%5D%3C%2Fp%3E%3Cp%3E%3Cem%3E%3Cstrong%3EThe%20final%20number%20of%20moles%20of%3A%3C%2Fstrong%3E%3C%2Fem%3E%3C%2Fp%3E%3Cul%3E%3Cli%3E%3Cem%3E%3Cstrong%3E%5Btex%5DA%20%3D%201-2%5Calpha%3D1-2%2A0.3%3D0.4mol%5Btex%5D%20%3C%2Fstrong%3E%3C%2Fem%3E%3C%2Fli%3E%3C%2Ful%3E%3Cul%3E%3Cli%3E%3Cem%3E%3Cstrong%3E%5Btex%5DB%3D2%281-%5Calpha%29%3D2%281-0.3%29%3D1.4mol%5Btex%5D%3C%2Fstrong%3E%3C%2Fem%3E%3C%2Fli%3E%3C%2Ful%3E%3Cul%3E%3Cli%3E%3Cem%3E%3Cstrong%3E%5Btex%5DD%3D1%2B2%5Calpha%3D1%2B2%2A0.3%3D1.6mol%5Btex%5D%3C%2Fstrong%3E%3C%2Fem%3E%3C%2Fli%3E%3C%2Ful%3E%3Cp%3EThus%20%2C%20total%20number%20of%20moles%20are%20%3A%200.4%2B1.4%2B0.9%2B1.6%3D4.3%3C%2Fp%3E%3Cp%3E%3Cstrong%3E%28i%29The%20mole%20fractions%20are%20%3A%20%3C%2Fstrong%3E%3C%2Fp%3E%3Cul%3E%3Cli%3E%3Cstrong%3E%5Btex%5DA%3D%5Cfrac%7B0.4%7D%7B4.3%7D%20%5C%5C%3D0.0930)
(ii)

Where ,
are the partial pressures of A,B,C,D respectively.
Total pressure = 1 bar .
∴
<em>
</em>
<em>
</em>
<em>
</em>
<em>
</em>

(iii)
Δ
ΔG = 
Answer:

Explanation:
Hello,
In this case, since the rms speed of the molecules is computed by:

Whereas the absolute temperature is computed from the internal energy (by using the Cp of helium (3.1156 J/g*K) as shown below:

Thereby, the rms speed results:

Regards.
Answer:
T₂ = 19.95°C
Explanation:
From the law of conservation of energy:

where,
mc = mass of copper = 37.2 g
Cc = specific heat of copper = 0.385 J/g.°C
mw = mass of water = 188 g
Cw = specific heat of water = 4.184 J/g.°C
ΔTc = Change in temperature of copper = 99.8°C - T₂
ΔTw = Change in temperature of water = T₂ - 18.5°C
T₂ = Final Temperature at Equilibrium = ?
Therefore,

<u>T₂ = 19.95°C</u>
<span>Intermolecular Forces present in HCl:
The Electronegativity difference between Chlorine and Hydrogen is 0.96 showing that the bond is polar covalent in nature. The Hydrogen atom is partially positive and Chlorine is partially positive making the molecule Dipole. Hence, the Intermolecular forces present in HCl are Dipole-Dipole Interactions.
</span>Intermolecular Forces present in CH₄:
The Electronegativity difference between Chlorine and Hydrogen is 0.35 showing that the bond is non-polar covalent in nature. Hence, the Intermolecular forces present in CH₄ are London Dispersion Forces.