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BARSIC [14]
1 year ago
13

A colored dye compound decomposes to give a colorless product. The original dye absorbs at 608 nm and has an extinction coeffici

ent of 4.7 ×10⁴ M⁻¹cm⁻¹ at that wavelength.
You perform the decomposition reaction in a 1-cm cuvette in a spectrometer and obtain the following data:
Time (min) - Absorbance at 608 nm
0 - 1.254
30 - 0.941
60 - 0.752
90 - 0.672
120 - 0.545
From these data, determine the rate law for the reaction "dye → product" and determine the rate constant.

Chemistry
1 answer:
dusya [7]1 year ago
6 0

Answer:

First Order reaction

k=  0.0067

Explanation:

To determine the rate law we will have to study the dependence of the rate of the reaction with respect to time.

So we need to calculate the concentrations given the data since  A= ε c l

where ε = molar absorptivity, c is the concentration moles/L, and l is the length of the cuvette.

c = A/εl

Time, min    M, (mol/L) x 10⁻⁵

0                         2.67

30                       2.00

60                       1.60

90                       1.43

120                      1.16

If the reaction were first order, the rate will have to remain constant which is not the case here since the rate of dissaperance is chnging:

Δmol/L/ΔT = (2.00 - 2.67 ) mol /30 min = -0.023 mol/Lmin ( between t= 30 and 0 min)

Δmol/L/ΔT = ( 0.160 - 2) /30 min =  = -0.061 mol/Lmin

If the rate is first order we will have to use the integrated rate law:

ln (A) t  = -kt + ln (A)₀

where (A) is the concentration, t is the time and k is the rate constant

Notice the equation  is of the form y = mx + b where

m= rate constant and b  (A)₀

To verify if our reaction is first order, we will graph the graph the data and see if it is a straght line and and if it is compute the slope which is k.

We could also perform a linear regression with an apprpiat program and obtain the results.

Working in excel the following results were obtained:

y = -0.0067 x  -10.582

k = 0.0067/ min

R² = correlation coefficient = 0.979

First Order reaction

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OLga [1]

Answer: 178.9 g

Explanation:

Density = \frac{mass}{volume in mL}

find volume of the cube: (5.80 cm) (5.80 cm) (5.80cm) = 195.112 cm³

1.0 cm³ = 1.0 mL

so 195.112 cm³ = 195.112 mL

plug value into density equation:

0.917 g/mL = (mass) / (195.112 mL)

and solve for mass!

3 0
1 year ago
For a particular reaction, the change in enthalpy is â9kJmole and the activation energy is 13kJmole. The enthalpy change (ÎH) an
Svetllana [295]

Answer:

The answer is Option a, that is "−9kJmole,5kJmole".

Explanation:

Please find the complete question in the attached file.

In the question, it uses the catalyst inside a process, which does not modify the process eigenvalues, however, it decreases the active energy with an enthalpy of -9kJmole, and also the power for activating decreases around 13 to 5 kJ mole, that's why the choice a is correct.

4 0
2 years ago
HA and HB are two strong monobasic acids. 25.0cm3 of 6.0mol/dm3 HA is mixed with 45.0cm3 of 3.0mol/dm3 HB.
Lostsunrise [7]

Answer:

The H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

(Option C)

Explanation:

Given;

concentration of HA, C_A = 6.0mol/dm³

volume of HA, V_A  = 25.0cm³, = 0.025dm³

Concentration of HB, C_B = 3.0mol/dm³

volume of HB, V_B = 45.0cm³ = 0.045dm³

To determine the H+ (aq) concentration in mol/dm³ in the resulting solution, we apply concentration formula;

C_iVi = C_fV_f

where;

C_i is initial concentration

V_i is initial volume

C_f is final concentration of the solution

V_f is final volume of the solution

C_iV_i = C_fV_f\\\\Based \ on \ this\ question, we \ can \ apply\ the \ formula\ as;\\\\C_A_iV_A_i + C_B_iV_B_i = C_fV_f\\\\C_A_iV_A_i + C_B_iV_B_i = C_f(V_A_i\ +V_B_i)\\\\6*0.025 \ + 3*0.045 = C_f(0.025 + 0.045)\\\\0.285 = C_f(0.07)\\\\C_f = \frac{0.285}{0.07} = 4.07 = 4.1 \ mol/dm^3

Therefore, the H+ (aq) concentration of the resulting solution is 4.1 mol/dm³

7 0
2 years ago
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igor_vitrenko [27]

Answer:

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When the potatoes were boiled it will surely produce no bubbles because the heat would have degrade the enzyme - catalase While the potatoes at room temperature potato produced the most bubbles because catalase works best at a room temperature.

If the potato solution was boiled for 10 minutes and cooled for 10 minutes before being tested, the average time for the disks to float to the surface of the hydrogen peroxide solution would be MORE THAN 30 SECONDS

5 0
2 years ago
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Answer:The amount of ice is equal to the amount of water.

Explanation:

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