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Burka [1]
3 years ago
9

Choose a slideshow or presentation program to organize your research. Your presentation should contain Information collected In

Part 1 and Part 2 Part 1: Include important facts found through your research. Include the answers to questions 1-4 for each of the three occupations you chose. Part 2: Include Information a visual representation of the data and the answers to questions 1-6. Submission Requirements: The student should submit a presentation created with a slideshow or presentation program of their choice. The presentation should include important facts obtained through research, references, Images, data, graphs, and answers to the reflection questions. Upload your presentation in the box below.​
Chemistry
1 answer:
beks73 [17]3 years ago
3 0

Answer:

Do the work

Explanation:

I know

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The following data were obtained in a kinetics study of the hypothetical reaction A + B + C → products. [A]0 (M) [B]0 (M) [C]0 (
Vladimir [108]

Answer:

B. First order, Order with respect to C = 1

Explanation:

The given kinetic data is as follows:

A + B + C → Products

     [A]₀     [B]₀    [C]₀       Initial Rate (10⁻³ M/s)

1.   0.4      0.4     0.2       160

2.  0.2      0.4      0.4       80

3.   0.6     0.1       0.2       15

4.   0.2     0.1       0.2        5

5.   0.2     0.2      0.4       20

The rate of the above reaction is given as:

Rate = k[A]^{x}[B]^{y}[C]^{z}

where x, y and z are the order with respect to A, B and C respectively.

k = rate constant

[A], [B], [C] are the concentrations

In the method of initial rates, the given reaction is run multiple times. The order with respect to a particular reactant is deduced by keeping the concentrations of the remaining reactants constant and measuring the rates. The ratio of the rates from the two runs gives the order relative to that reactant.

Order w.r.t A : Use trials 3 and 4

\frac{Rate3}{Rate4}= [\frac{[A(3)]}{[A(4)]}]^{x}

\frac{15}{5}= [\frac{[0.6]}{[0.2]}]^{x}

3 = 3^{x} \\\\x =1

Order w.r.t B : Use trials 2 and 5

\frac{Rate2}{Rate5}= [\frac{[B(2)]}{[B(5)]}]^{y}

\frac{80}{20}= [\frac{[0.4]}{[0.2]}]^{y}

4 = 2^{y} \\\\y =2

Order w.r.t C : Use trials 1 and 2

\frac{Rate1}{Rate2}= [\frac{[A(1)]}{[A(2)]}]^{x}[\frac{[B(1)]}{[B(2)]}]^{y}[\frac{[C(1)]}{[C(2)]}]^{z}

we know that x = 1 and y = 2, substituting the appropriate values in the above equation gives:

\frac{160}{80}= [\frac{[0.4]}{[0.2]}]^{1}[\frac{[0.4]}{[0.4]}]^{2}[\frac{[0.2]}{[0.4]}]^{z}

1 = (0.5)^{z}

z = 1

Therefore, order w.r.t C = 1

8 0
3 years ago
A sample of an unknown metal has a mass of 120.4 g. As the sample cools from 90.5°C to 25.7°C, it releases 7020 J of energy. Usi
Dafna1 [17]

Answer:

Explanation:

sheesh i wouldn’t know

6 0
3 years ago
___Mg + ___Fe2O3 -> ___MgO + ___Fe
kipiarov [429]
3 Mg, 0 Fe2O3, 3MgO, 2 Fe
5 0
3 years ago
The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is __________
PIT_PIT [208]

Molality is one way of expressing concentration of a solute in a solution. It is expressed as the mole of solute per kilogram of the solvent. To calculate for the molality of the given solution, we need to convert the mass of solute into moles and divide it to the mass of the solvent.

<span>
Moles of HCl = 5.5 g HCl ( 1 mol HCl / 36.46 g HCl ) = 0.1509 mol HCl</span>

<span>
Molality = 0.1509 mol HCl / 200 g C2H6O ( 1 kg / 1000 g )
Molality = 0.7543 mol / kg</span>


<span>The concentration in molality of hcl in a solution that is prepared by dissolving 5.5 g of hcl in 200.0 g of c2h6o is 0.7453 molal.</span>
6 0
3 years ago
In the laboratory, a general chemistry student measured the pH of a 0.529 M aqueous solution of phenol (a weak acid), C6H5OH to
Artyom0805 [142]

Answer:

The dissociation constant of phenol from given information is 9.34\times 10^{-11}.

Explanation:

The measured pH of the solution = 5.153

C_6H_5OH\rightarrow C_6H_5O^-+H^+

Initially      c

At eq'm   c-x       x  x

The expression of dissociation constant is given as:

K_a=\frac{[C_6H_5O^-][H^+]}{[C_6H_5OOH]}

Concentration of phenoxide ions and hydrogen ions are equal to x.

pH=-\log[x]

5.153=-\log[x]

x=7.03\times 10^{-6} M

K_a=\frac{x\times x}{(c-x)}=\frac{x^2}{(c-x)}=\frac{(7.03\times 10^{-6} M)^2}{ 0.529 M-7.03\times 10^{-6} M}

K_a=9.34\times 10^{-11}

The dissociation constant of phenol from given information is 9.34\times 10^{-11}.

4 0
3 years ago
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