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noname [10]
3 years ago
10

Are dienes, saturated or unsaturated molecules?

Chemistry
1 answer:
Lana71 [14]3 years ago
6 0

Unsaturated, as they have C=C bonds
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If the K a Ka of a monoprotic weak acid is 7.3 × 10 − 6 , 7.3×10−6, what is the pH pH of a 0.40 M 0.40 M solution of this acid?
olga_2 [115]

Answer:

pH =3.8

Explanation:

Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:

HA + H₂O   ⇄ H₃O⁺ + A⁻    with  Ka = [ H₃O⁺] x [A⁻]/ [HA]

The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.

In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:

                          HA                                   H₃O⁺                          A⁻          

Initial, M             0.40                                   0                              0

Change , M          -x                                     +x                            +x

Equilibrium, M    0.40 - x                              x                               x

Lets express these concentrations in terms of the equilibrium constant:

Ka = x² / (0.40 - x )

Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,

7.3 x 10⁻⁶ = x² / 0.40  ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³

[H₃O⁺] = 1.71 x 10⁻³

Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.

pH = - log ( 1.71 x 10⁻³ ) = 3.8

Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.

4 0
3 years ago
How many mL of a 0.375 M solution can be made from 35 g of calcium phosphate?
andrew11 [14]

Answer:

300 mL

Explanation:

the unit formula of calcium phosphate is Ca3(PO4)2

molar mass of Ca3(PO4)2 = (3×40 + 2×31 + 8×16) g/mol = 310 g/mol

n = m/M = 35 g/(310 g/mol)

c = n/V

V = n/c = [35 g/(310 g/mol)]/0.375 mol/L

V = 0.30 L = 300 mL

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1.53 moles of Fe is your solution hope it helps!
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