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Vikki [24]
2 years ago
5

Use the Lab Report document as a template. Use the descriptions and examples of each part to help you write your Lab Report. You

r Lab Report should focus on the results of the experiment that you conducted, not on the results of the experiment provided with the Image Analyzer.Answer these questions in the introduction of your Lab Report:What is the principle behind the technique of paper chromatography?What are some uses of paper chromatography?What was the purpose of your experiment?Answer these questions in the discussion and conclusion of your Lab Report:What does your data tell you about the markers tested?What did you use for controls in your experiment?What are some sources of error in your experiment?How could you modify your experiment?Be sure to look at the Lab Report Rubric to see how you will be graded.
Chemistry
1 answer:
GarryVolchara [31]2 years ago
6 0
This link might help you!!

https://www.eiu.edu/biology/bio1500/writing_a_lab_report.pdf
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Help plzzzzzzz ASAP!
Agata [3.3K]
I am going to say C. it has to do with the angles
8 0
3 years ago
Read 2 more answers
A sample of PCl5(g) was placed in an otherwise empty flask at an initial pressure of 0.500 atm and a temperature above 500 K. Ov
tia_tia [17]

Answer:

Kp = 0.81666

Explanation:

Pressure of PCl₅ = 0.500 atm

Considering the ICE table for the equilibrium as:

                     PCl₅ (g)      ⇔          PCl₃ (g) +       Cl₂ (g)

t = o               0.500

t = eq                -x                             x                      x

--------------------------------------------- --------------------------

Moles at eq: 0.500-x                       x                      x

       

Given the pressure of PCl₅ at equilibrium = 0.150 atm

Thus, 0.500 - x = 0.150

x = 0.350 atm

The expression for the equilibrium constant is:

K_p=\frac {P_{PCl_3}P_{[Cl_2}}{P_{PCl_5}}  

So,

K_p=\frac{x^2}{0.500-x}  

x = 0.350 atm

Thus,

K_p=\frac{{0.350}^2}{0.500-0.350}  

<u>Thus, Kp = 0.81666</u>

7 0
3 years ago
El agua del mar contiene aproximadamente un 3,0 % m/v de sal (NaCl, 58,44 g/mol), (asuma que es la única fuente de cloruros) si
Alchen [17]

Answer:

s = 4.41 g/L.

Explanation:

¡Hola!

En este caso, considerando el escenario dado, se hace necesario para nosotros saber que la posible reacción de disociación la experimenta el cloruro de plomo (II) como se muestra a continuación:

PbCl_2(s)\rightleftharpoons 2Cl^-(aq)+Pb^{2+}(aq)

Lo cual hace que la expresión de equilibrio se calcule como:

Ksp=[Pb^{2+}][Cl^-]^2

Y que en términos de la solubilidad molar, s, se resuelve como:

1.6x10^{-5}=s(2s)^2\\\\1.6x10^{-5}=4s^3\\\\s=\sqrt[3]{\frac{1.6x10^{-5}}{4} } \\\\s=0.0159molPbCl_2/L

Ahora, convertimos este valor a g/L al multiplicarlo por la masa molar del cloruro de plomo (II):

s=0.0159molPbCl_2/L*\frac{278.1gmolPbCl_2}{1molmolPbCl_2} \\\\s=4.41g/L

¡Saludos!

7 0
3 years ago
Four students give their teacher identical apples. Each student sets his or her apple on a different stack of books. Which
ycow [4]
D
explanation ; i’m smart
7 0
3 years ago
Student perfotms a Benedict's test on an unknown substance. He adds reagent(the chemical required to make a color change), and n
elixir [45]

Answer:

Reducing sugars are absent

Explanation:

Benedict's solution is an substance used in testing sugars. It is mixture of sodium carbonate, sodium citrate and copper(II) sulfate pentahydrate. It can be used instead of Fehling's solution in testing for the presence of reducing sugars.

Reducing sugars contain the -CHO group. If there is no colour change after the addition of Benedict's solution, then we can conclude that reducing sugars are absent.

7 0
3 years ago
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