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dsp73
3 years ago
12

3. the [OH-] of a solution with a [H+] of 9.3 x 10^-4 M

Chemistry
1 answer:
Ostrovityanka [42]3 years ago
3 0

Answer:

[OH-] = 1.08x10^-11 M

Explanation:

From the question given,

Concentration of Hydrogen ion, [H+]

= 9.3x10^-4 M

Concentration of Hydroxide ion, [OH-] =..?

The concentration of the hydrogen ion, [H+] and hydroxide ion, [OH-] are related with the following formula:

[H+] x [OH-] = 1x10^-14

Using the above formula, we can easily calculate the concentration of the hydroxide ion, [OH-] as follow:

[H+] x [OH-] = 1x10^-14

[H+] = 9.3x10^-4 M

9.3x10^-4 x [OH-] = 1x10^-14

Divide both side by 9.3x10^-4

[OH-] = 1x10^-14 /9.3x10^-4

[OH-] = 1.08x10^-11 M

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tiny-mole [99]

Answer:

The H in the carboxyl group.

Explanation:

Acetic acid can be written as CH₃COOH, where -COOH is the functional group carboxyl, responsible for the acidity of organic acids. The H in the carboxyl group is the one that is donated in the acid reaction.

CH₃COOH(aq) + H₂O(l) ⇄ CH₃COO⁻(aq) + H₃O⁺

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3 years ago
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Because the density of water is one

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3 years ago
COCI2 has an effusion rate of 0.00172 m/sec. Which of the gases below would have an effusion rate of 0.00323 m/sec?
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2 years ago
Consider the decomposition of a metal oxide to its elements, where M represents a generic metal. M 3 O 4 ( s ) − ⇀ ↽ − 3 M ( s )
Citrus2011 [14]

Answer:

a) ΔGrxn = 6.7 kJ/mol

b) K = 0.066

c) PO2 = 0.16 atm

Explanation:

a) The reaction is:

M₂O₃ = 2M + 3/2O₂

The expression for Gibbs energy is:

ΔGrxn = ∑Gproducts - ∑Greactants

Where

M₂O₃ = -6.7 kJ/mol

M = 0

O₂ = 0

deltaG_{rxn} =((2*0)+(3/2*0))-(1*(-6.7))=6.7kJ/mol

b) To calculate the constant we have the following expression:

lnK=-\frac{deltaG_{rxn} }{RT}

Where

ΔGrxn = 6.7 kJ/mol = 6700 J/mol

T = 298 K

R = 8.314 J/mol K

lnK=-\frac{6700}{8.314*298} =-2.704\\K=0.066

c) The equilibrium pressure of O₂ over M is:

K=P_{O2} ^{3/2} \\P_{O2}=K^{2/3} =0.066^{2/3} =0.16atm

3 0
3 years ago
PH is 7.45. Calculate value of [H3O+] and [OH-]
lana66690 [7]

Answer:

The answer is (H30+) =3,55e-8M and (OH-)=2,82e-7M

Explanation:

We use the formulas:

pH= - log(H30+)  and Kwater=(H30+)x(OH-)

pH= - log(H30+)  ----< (H30+)= antilog- pH=antilog- 7,45=3,55E-8M

Kwater=(H30+)x(OH-)

(OH-)=Kwater/(H30+)= 1,00e-14/3,55e-8 = 2,82e-7

8 0
3 years ago
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