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denis23 [38]
3 years ago
13

For the diprotic weak acid h2a, ka1 = 3.2 × 10-6 and ka2 = 6.1 × 10-9. what is the ph of a 0.0650 m solution of h2a? what are th

e equilibrium concentrations of h2a and a2– in this solution?
Chemistry
1 answer:
Stolb23 [73]3 years ago
5 0
Given:

Diprotic weak acid H2A:
 
Ka1 = 3.2 x 10^-6
Ka2 = 6.1 x 10^-9. 
Concentration = 0.0650 m 

Balanced chemical equation:

H2A ===> 2H+  + A2- 
0.0650       0        0
-x                2x       x
------------------------------
0.065 - x     2x      x

ka1 = 3.2 x 10^-6 = [2x]^2 * [x] / (0.065 - x)

solve for x and determine the concentration at equilibrium. 


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suppose medical records indicate that the length of newborn babies (in inches) is normally distributed with a mean of 20 and a s
Tanya [424]

Answer:

  • <u>0.6826 (68.26%)</u>

Explanation:

<u>1) Find the z-scores:</u>

  • z = [X - μ ] / σ

a) z-score for 22.6 inches length

  • X = 22.6
  • μ = 20
  • σ = 2.6

  • z = [ 22.6 - 20 ] / 2.6 = 1.00

b) z-score for 17.4 inches length

  • X = 17.4
  • μ = 20
  • σ = 2.6

  • z = [ 17.4 - 20 ] / 2.6 =  - 1.00

<u>2) Probability</u>

Then, you have to find the probability that the length of an infant is between - 1.00 and 1.00 standards deviations (σ) from the mean (μ).

That is a well known value of 68%, which is part of the 68-95-99.7 empirical rule.

The most exact result is obtained from tables and is 68.26%:

  • 1 - P (z ≥ 1.00) - P (z ≤ - 1.00) = 1 - 0.1587 - 0.1587 = 0.6826 = 68.26%
5 0
3 years ago
The following data were collected for the rate of disappearance of NO in the reaction 2NO(g)+O2(g)→2NO2(g)::
Anit [1.1K]

Answer:

a) The rate law is: v = k[NO]² [O₂]

b) The units are: M⁻² s⁻¹

c) The average value of the constant is: 7.11 x 10³ M⁻² s⁻¹

d) The rate of disappearance of NO is 0.8 M/s

e) The rate of disappearance of O₂ is 0.4 M/s

Explanation:

The experimental rates obtained can be expressed as follows:

v1 = k ([NO]₁)ᵃ ([O₂]₁)ᵇ = 1.41 x 10⁻² M/s

v2 = k ([NO]₂)ᵃ ([O₂]₂)ᵇ = 5.64 x 10⁻² M/s

v3 = k ([NO]₃)ᵃ ([O₂]₃)ᵇ = 1.13 x 10⁻¹ M/s

where:

k = rate constant

[NO]₁ = concentration of NO in experiment 1

[NO]₂ = concentration of NO in experiment 2

[NO]₃ = concentration of NO in experiment 3

[O₂]₁ = concentration of O₂ in experiment 1

[O₂]₂ = concentration of O₂ in experiment 2

[O₂]₃ = concentration of O₂ in experiment 3

a and b = order of the reaction for each reactive respectively.

We can see these equivalences:

[NO]₂ = 2[NO]₁

[O₂]₂ = [O₂]₁

[NO]₃ = [NO]₂

[O₂]₃ = 2[O₂]₂

So, v2 can be written in terms of the concentrations used in experiment 1 replacing [NO]₂ for 2[NO]₁ and [O₂]₂ by [O₂]₁ :

v2 = k (2 [NO]₁)ᵃ ([O₂]₁)ᵇ

If we rationalize v2/v1, we will have:

v2/v1 = k *2ᵃ * ([NO]₁)ᵃ * ([O₂]₁)ᵇ / k * ([NO]₁)ᵃ * ([O₂]₁)ᵇ (the exponent "a" has been distributed)

v2/v1 = 2ᵃ

ln(v2/v1) = a ln2

ln(v2/v1) / ln 2 = a

a = 2

(Please review the logarithmic properties if neccesary)

In the same way, we can find b using the data from experiment 2 and 3 and writting v3 in terms of the concentrations used in experiment 2:

v3/v2 = k ([NO]₂)² * 2ᵇ * ([O₂]₁)ᵇ / k * ([NO]₂)² * ([O₂]₂)ᵇ

v3/v2 = 2ᵇ

ln(v3/v2) = b ln 2

ln(v3/v2) / ln 2 = b

b = 1

Then, the rate law for the reaction is:

<u>v = k[NO]² [O₂]</u>

Since the unit of v is M/s and the product of the concentrations will give a unit of M³, the units of k are:

M/s = k * M³

M/s * M⁻³ = k

<u>M⁻² s⁻¹ = k </u>

To obtain the value of k, we can solve this equation for every experiment:

k = v / [NO]² [O₂]

for experiment 1:

k = 1.41 x 10⁻² M/s / (0.0126 M)² * 0.0125 M = 7.11 x 10³ M⁻² s⁻¹

for experiment 2:

k = 7.11 x 10³ M⁻² s⁻¹

for experiment 3:

k = 7.12 x 10³ M⁻² s⁻¹

The average value of k is then:

(7.11 + 7.11 + 7.12) x 10³ M⁻² s⁻¹ / 3 = <u>7.11 x 10³ M⁻² s⁻¹ </u>

The rate of the reaction when [NO] = 0.0750 M and [O2] =0.0100 M is:

v = k [NO]² [O₂]

The rate of the reaction in terms of the disappearance of NO can be written this way:

v = 1/2(Δ [NO] / Δt) (it is divided by 2 because of the stoichiometric coefficient of NO)

where (Δ [NO] / Δt) is the rate of disappearance of NO.

Then, calculating v with the data provided by the problem:

v = 7.11 x 10³ M⁻² s⁻¹ * (0.0750M)² * 0.0100M = 0.4 M/s

Then, the rate of disappearance of NO will be:

2v = Δ [NO] / Δt = <u>0.8 M/s</u>

The rate of disappearance of O₂ has to be half the rate of disappearance of NO because two moles of NO react with one of O₂. Then Δ [O₂] / Δt = <u>0.4 M/s</u>

With calculations:

v = Δ [O₂] / Δt = 0.4 M/s (since the stoichiometric coefficient is 1, the rate of disappearance of O₂ equals the rate of the reaction).

3 0
3 years ago
Determine whether each of the properties described applies to volumetric or graduated glassware. 1. Used for applications in whi
kogti [31]

Answer:

1) volumetric

2) graduated

3) volumetric

Explanation:

A volumetric glassware is a glassware that is marked at a particular point. A typical example of a volumetric glassware is the volumetric flask. A volumetric glassware is capable of measuring only a specific volume of a liquid.

On the other hand, graduated glassware can measure a range of volumes of liquid. However, a volumetric glassware is still required where a high degree of accuracy is important.

5 0
2 years ago
Which is bigger 145 m or 145 km
Citrus2011 [14]
145 km i think is bigger
8 0
3 years ago
Read 2 more answers
Is matter smooth and continuous?
Strike441 [17]

Answer:

<h2><u><em>YES,</em></u></h2>

Explanation:

<h2><u><em> Matter is smooth and continuous. ... A liquid has a definite volume; but it has no definite shape.</em></u></h2>
3 0
3 years ago
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