F = ma = (kg)(m/s2) = kg ´ m/s2 N
hope this helps :D
Answer:
Lithium hydroxide is a base.
Carbon dioxide is the anhydride of the carbonic acid, H₂CO₃.
Therefore, the reaction awaited is a typical neutralization reaction with the formation of a salt and water.
2LiOH + CO₂ → Li₂CO₃ + H₂O
So, 2*20 = 40 moles of LiOH react with 20 moles of CO₂.
Molar Mass of LiOH = 23.95 g/mol
So, 40 * 23.95 = 958 g
Use higher concentration of acid to produce 2 moles of hydrogen to form water
Answer:
See explanation.
Explanation:
Since potassium tert-butoxide is a strong bulky base, we expect that an elimination reaction predominates.
This reaction is expected to occur by E2 mechanism. The cis isomer of 1-bromo-4-tert-butylcyclohexane is known to react faster by E2 mechanism because it fulfills the anticoplanar arrangement required for for E2 mechanism, where the leaving group and adjacent proton must be anti to each other and in the same plane. The trans isomer can not fulfill this requirement.
Answer:
(i)The mole fractions are :
(ii)![K=0.4508](https://tex.z-dn.net/?f=K%3D0.4508)
(iii)ΔG = 1.974kJ
Explanation:
The given equation is :
⇄![3C+2D](https://tex.z-dn.net/?f=3C%2B2D)
Let
be the number of moles dissociated per mole of ![B](https://tex.z-dn.net/?f=B)
Thus ,
<em>The initial number of moles of :</em>
+
⇄
+ ![2D\\(1+2\alpha)](https://tex.z-dn.net/?f=2D%5C%5C%281%2B2%5Calpha%29)
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)
![K=\frac{(P_C^3)(P_D^2)}{(P_A^2)(P_B)}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%28P_C%5E3%29%28P_D%5E2%29%7D%7B%28P_A%5E2%29%28P_B%29%7D)
Where ,
are the partial pressures of A,B,C,D respectively.
Total pressure = 1 bar .
∴
<em>
</em>
<em>
</em>
<em>
</em>
<em>
</em>
![K=\frac{0.2093^3*0.3721^2}{0.0930^2*0.3256} \\K=0.4508](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B0.2093%5E3%2A0.3721%5E2%7D%7B0.0930%5E2%2A0.3256%7D%20%5C%5CK%3D0.4508)
(iii)
Δ![G=-RTlnK\\](https://tex.z-dn.net/?f=G%3D-RTlnK%5C%5C)
ΔG = ![-8.314*(273+25)*ln(0.4508)\\=1973.96J\\=1.974kJ](https://tex.z-dn.net/?f=-8.314%2A%28273%2B25%29%2Aln%280.4508%29%5C%5C%3D1973.96J%5C%5C%3D1.974kJ)