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Whitepunk [10]
2 years ago
10

A nonpolar solvent has the highest solubility with a(n) _____ solute. A. polar B. nonpolar C. ionic D. electrically charged solu

te. ​

Chemistry
1 answer:
jenyasd209 [6]2 years ago
6 0

Answer:

nonpolar

Explanation:

A nonpolar solvent has the highest solubility with a <u>nonpolar</u> solute.

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H2SO4 is a strong acid because the first proton ionizes 100%. The Ka of the second proton is 1.1x10-2. What would be the pH of a
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Question options:

a) 2.05

b) 0.963

c) 0.955

d) 1.00

Answer:

b) 0.963

Explanation:

H2SO4→ HSO4- + H3O+

HSO4- + H2O ⇌ SO42- + H3O+

Construct ICE table:

        HSO4- (aq)    +    H2O        ⇌      SO42- (aq)     +     H3O+ (aq)

I          0.1                  solid &                   0                          0.1

C         -x                     liquid                 + x                            + x

E         0.1 - x          are ignored              x                          0.1 + x

Calculate x

Ka = products/reactants

  = \frac{[SO42-] [H3O+]}{[HSO4-]}

0.011 = \frac{x (0.1 + x)}{0.1 - x}

0.011 x (0.1 -x) = o.1x + x^2

0.0011 - 0.011 x - o.1x - x^2 = 0

0.0011 - 0.011 x - x^2 = 0

Use formula to solve for quadratic equation

x = { -b +,-\sqrt{b^2 - 4ac / 2a

a = -1, b = -0.111, c = 0.001

Solve for x

x = \sqrt[-(-o.111)]{(-0.111)^2 - 4(-1) (0.0011) }  / 2(-1)

x = 0.111 +,- \sqrt{0.012321 + 0.0044} / -2

x = 0.111 +,- \sqrt{0.016721} / -2

x = \frac{0.111 +, - 0.1293}{-2}

x = \frac{0.111 + 0.1293}{-2}   , x = \frac{0.111  - 0.1293}{-2}

x = \frac{0.2403}{-2}    , x = \frac{0.0183}{-2}

x = - 0.12015  , x = 0.00915

x cannot be negative, so

x = 0.00915 M

Calculate [H3O+]

[H3O+] = 0.1 M + x

[H3O+] = 0.1 M + 0.00915 M

[H3O+] = 0.10915 M

Clculate pH

pH = - log [ H3O+]

pH = - log [ 0.10915]

pH = 0.963

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He resonance structures of carbon monoxide are shown below. show how each structure can be converted into the other using the cu
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The two resonating structures of Carbon Monoxide are shown below.
The movement of electrons are shown by arrow.

Structure A:
                  In structure a A the formal charges of Carbon and Oxygen are zero. As Formal Charge is calculated as,

             = # of valence electron - electrons in lone pairs + 1/2 bonding                            electrons electrons

For C:
           = 4 - 2 + 4/2
           = 4- 4
           = 0

For O:
           = 6 - 4 + 4/2
           = 6- 6
           = 0

Structure B:

Formal Charge on C:

           = 4 - 2 + 6/2
           = 4- 5
           = -1

Formal Charge on O:

           = 6 - 2 + 6/2
           = 6 - 5
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Molecular geometry of HCO2-
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Absolutely

Explanation:

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3 years ago
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