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Vinil7 [7]
3 years ago
6

Does anyone have the data table answers for 8.03 - Solutions Lab Report?

Chemistry
1 answer:
choli [55]3 years ago
3 0

A laboratory report is an important part of the scientific process that communicate the important work that has been done.

<h3>How to depict the laboratory report?</h3>

Your information is incomplete. Therefore, an overview of a laboratory report will be given. The laboratory report is important for future studies a d experiments.

A laboratory report is broken down intosections such as title, abstract, introduction, methods, materials, results, discussion conclusion, and references.

Learn more about solutions lab report on:

brainly.com/question/26974096

#SPJ1

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Deuterium (D or 2H) is an isotope of hydrogen. The molecule D2 undergoes an exchange reaction with ordinary hydrogen, H2, that l
Paladinen [302]

Answer:

Kp_2} =2.0

Explanation:

K_{p1}= 1.8

K_{p2}= ???

T_1= 298K

T_2 = 415 K

\delta H = 0.64 kJ/mol = 640 J/mol

R = 8.314 J/mol.K

Using Van't Hoff Equation:

In (\frac{Kp_2}{Kp_1} )=(-\frac{DH}{R} *(\frac{1}{T_2}-\frac{1}{T_1}  )

In (\frac{Kp_2}{1.8} )=(-\frac{640}{8.314} *(\frac{1}{415}-\frac{1}{298}  )

In (\frac{Kp_2}{1.8} )=0.072827

(\frac{Kp_2}{1.8} )= e^{0.0728727}

\frac{Kp_2}{1.8} =1.075593605

Kp_2} =1.075593605*1.8

Kp_2} =1.936068489

Kp_2} =2.0

3 0
3 years ago
Read 2 more answers
Do all multicellular organisms follow the same pattern of organization?
natta225 [31]
So so sorry if i'm wrong but i'm pretty sure no
3 0
4 years ago
Using the periodic table to locate each element, write the electron configuration of(c) Re.
Aleksandr [31]

Rhenium is a chemical element with the symbol Re and atomic number 75. The electron configuration of Re is [Xe] 6s^{2} 4f^{14} 5d^{5}.

<h3>How to write an electronic configuration?</h3>

1. Identify the given element and its atomic number from the periodic table.

2. Write the electron configuration by the energy level and the type of orbital first, then the number of electrons present in the orbital as superscript.

The easiest way to write the electronic configuration for any element is by   using a diagonal rule for electron filling order in the different subshells according to the Aufbau principle.

The 3 rules for writing the electron configuration in the orbital box diagram are – the Aufbau rule, the Pauli-exclusion rule, and Hund's Rule.

To learn more about electronic configuration, refer

https://brainly.ph/question/73419

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5 0
2 years ago
At a certain temperature the vapor pressure of pure heptane (C_7 H_16) is measured to be 170. torr. Suppose a solution is prepar
steposvetlana [31]

Explanation:

The given data is as follows.

       Pressure (P) = 170 torr,      mass of heptane (m) = 86.7 g

First, we will calculate the number of moles as follows.

          No. of moles = \frac{mass}{\text{molar mass}}

                        = \frac{86.7 g}{100 g/mol}

                        = 0.867 mol

Now, the number of moles of CH_{3}COBr are calculated as follows.

     No. of moles =  \frac{mass}{\text{molar mass}}

                           =  \frac{125}{122.9}    

                           = 1.07

Therefore, mole fraction of heptane will be calculated as follows.

     Mole fraction = \frac{\text{moles of heptane}}{\text{total moles}}

                           = \frac{0.867}{0.867 + 1.07}

                           = \frac{0.867}{1.937}

                           = 0.445

Now, we will calculate the partial pressure of heptane as follows.

            P_{A} = x_{A}P^{o}_{A}

                 = 170 \times 0.445      

                 = 75.65 torr

Thus, we can conclude that the partial pressure of heptane vapor above this solution is 75.65 torr.

8 0
4 years ago
Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial conc
Anni [7]

Answer:

Ka = 4.76108

Explanation:

  • CO(g) + 2H2(g) ↔ CH3OH(g)

∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]

                      [ ]initial         change         [ ]eq

CO(g)              0.27 M       0.27 - x        0.27 - x

H2(g)              0.49 M       0.49 - x        0.49 - x

CH3OH(g)          0                0 + x               x = 0.11 M

replacing in Ka:

⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)

⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)

⇒ Ka = (0.11) / (0.38)²(0.16)

⇒ Ka = 4.76108

7 0
3 years ago
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