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a_sh-v [17]
3 years ago
10

Which substance do you expect to be most soluble in water? Group of answer choices Methane, CH4 Nitrogen, N2 Ammonia, NH3 Carbon

dioxide, CO2
Chemistry
1 answer:
sdas [7]3 years ago
6 0

Answer:

Ammonia, NH3

Explanation:

The water is a polar solvent, which means that it has a dipole moment different from 0. The polar solvents solubles polar solutes.

The polarity can be found by the VSPER theory, where the bonds and the lone pairs of electrons at the central atom are the most distance possible. If the bonds are done by elements with close electronegativity, it's nonpolar, if it's done with elements with different electronegativities, it's polar.

The polar moment of the bonds are vectors if the sum of it is 0, the molecule is nonpolar, if it's different from 0, the molecule is polar.

Methane, CH4, has a tetrahedral geometry, and the bonds of C-H are nonpolar, thus it is a nonpolar molecule.

Nitrogen, N2, has a linear geometry and is nonpolar (bonds with

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The 3 in 4p3 represents:
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In the gas-phase reaction 2 A + B ⇌ 3 C + 2 D, it was found that, when 1.00 mol A, 2.00 mol B, and 1.00 mol D were mixed and all
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Answer:

(i)The mole fractions are :

  • A=\frac{0.4}{4.3} \\=0.0930
  • B=\frac{1.4}{4.3} \\=0.3256
  • C=\frac{0.9}{4.3} \\=0.2093
  • D=\frac{1.6}{4.3} \\=0.3721

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(iii)ΔG = 1.974kJ

Explanation:

The given equation is :

2A+B⇄3C+2D

Let \alpha be the number of moles dissociated per mole of B

Thus ,

<em>The initial number of moles of :</em>

  • A=1
  • B=2
  • C=0
  • D=1

2A\\(1-2\alpha)     +  B\\2(1-\alpha)  ⇄  3C\\(3\alpha)   + 2D\\(1+2\alpha)

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

  • B=\frac{1.4}{4.3} \\=0.3256
  • C=\frac{0.9}{4.3} \\=0.2093
  • D=\frac{1.6}{4.3} \\=0.3721

(ii)

K=\frac{(P_C^3)(P_D^2)}{(P_A^2)(P_B)}

Where ,

P_A,P_B,P_C,P_D are the partial pressures of A,B,C,D respectively.

Total pressure = 1 bar .

∴

<em>P_A= 0.0930*1=0.0930</em>

<em>P_B= 0.3256*1=0.3256</em>

<em>P_C= 0.2093*1=0.2093</em>

<em>P_D= 0.3721*1=0.3721</em>

K=\frac{0.2093^3*0.3721^2}{0.0930^2*0.3256} \\K=0.4508

(iii)

ΔG=-RTlnK\\

ΔG = -8.314*(273+25)*ln(0.4508)\\=1973.96J\\=1.974kJ

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Explanation:

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