Answer:
5.6 L
Explanation:
We can apply Charles' Law here since our pressure is constant (will not change inside the refrigerator) and we are relating change in volume with change in temperature:
V₁ / T₁ = V₂ / T₂ where V₁ and T₁ are initial volume and temperature, and V₂ and T₂ are final volume and temperature. Let's plug in what we know and solve for the unknown:
28.0 L / 25 °C = V₂ / 5 °C => V₂ = 5.6 L
5.6 L is our new volume (at 5 °C).
Your answer would be 172.1703 g/ml
U got kik?
lets talk first before we go any further
75% would be your answer since the lower letter represents a shorter plant
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 = 