I'd say it's suspension, but I'm not 100% sure...
Answer:
1 and 3
Explanation:
The vertical columns (groups) of the periodic table are arranged such that all its elements have the same number of valence electrons. All elements within a certain group thus share similar properties.
0! because you walked back in forth in diferent direcions
<span>70.4 mg CO2 x 1.0 g /1000 mg x 1 mole CO2/ 44 gCO2 x 1 mole C/1 mole CO2 = 0.0016 moles C
14.4 mg H2O x 1.0 g/1000 mg x 1 mole H2O/18 g H2O x 2 moles H/ 1 mole H2O = 0.0016 moles O
molar mass of C=12 g/mole
molar mass of H=1 g/mole
0.0016 moles C x 12 g C/ 1 mole C = 0.0192 g C or 19.2 mg C
0.00156 moles H x 1 g H/1 mole H = 0.00156 g H or 1.56 mg H
mg O= 30.4 mg vanillin - 19.2 mg C – 1.56 mg H = 9.64 mg O
molar mass of O=16 g/mole
9.64 mg O x 1 g/1000 mg x 1 mole O/16.0 g = 0.000602
C.0016 H.0016 O.000602; divide all the moles by the smallest value of0.000602
C2.66H2.66O1 is the empirical formula;
to obtain whole numbers multiply by 3
3[C2.66H2.66O1] = C8H8O3
above formula weight: 8(C) + 8(H) + 3(O) = 8(12) + 8(1) + 3(16) = 152 amu
The empirical formula weight and the molecular formula weight are the same .
Molecular formula is C8H8O3.</span>
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 = 