Answer:
Asking for and receiving a paper that someone who is in another lab section wrote is an honor violation. True
An oral discussion with a classmate regarding a paper's topics and format is cheating. False
Purchasing papers from tutoring companies is allowed. False
Sending a paper that you wrote to a student who is currently in another section is an honor violation. True
Misconduct in research or scholarship includes fabrication, falsification, or plagiarism in proposing, performing, reviewing, or reporting research. It does not include honest error or honest differences in interpretations or judgements of data. True
Discussing an exam or laboratory practical with anyone prior to the conclusion of the week that final exams and laboratory practicals are given would be considered an honor violation. True
Explanation:
The Aggie code of honour is a code of academic integrity of the Texas A&M University. It spells out the codes of academic integrity and responsible research. Instructors are to include this code in all syllabi. It targets the application of the highest degree of integrity in academic research and forbids misconducts such as cheating, plagiarism, falsification et cetera.
Answer:
The correct option is: b. pH 6.4-8.0
Explanation:
Phenol red is a weak acid that is used as a pH indicator and exists in the form of stable red crystals.
<u>The color of the phenol red solution changes from yellow to red when the change in pH is observed. The color of phenol red transitions from yellow to red when the pH is 6.8 - 8.2 or 6.4 - 8.0</u>
Above the pH of 8.2, the phenol red solution turns a bright pink in color.
It is used to 'date' samples as scientists can measure the amount of Carbon-14 isotope compared to a living organism.
<u>Answer:</u> The molar mass of unknown gas is 367.12 g/mol
<u>Explanation:</u>
Rate of a gas is defined as the amount of gas displaced in a given amount of time.
![\text{Rate}=\frac{V}{t}](https://tex.z-dn.net/?f=%5Ctext%7BRate%7D%3D%5Cfrac%7BV%7D%7Bt%7D)
To calculate the rate of diffusion of gas, we use Graham's Law.
This law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:
![\text{Rate of effusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}](https://tex.z-dn.net/?f=%5Ctext%7BRate%20of%20effusion%7D%5Cpropto%20%5Cfrac%7B1%7D%7B%5Csqrt%7B%5Ctext%7BMolar%20mass%20of%20the%20gas%7D%7D%7D)
So,
![\left(\frac{\frac{V_{X}}{t_{X}}}{\frac{V_{O_2}}{t_{O_2}}}\right)=\sqrt{\frac{M_{O_2}}{M_{X}}}](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7B%5Cfrac%7BV_%7BX%7D%7D%7Bt_%7BX%7D%7D%7D%7B%5Cfrac%7BV_%7BO_2%7D%7D%7Bt_%7BO_2%7D%7D%7D%5Cright%29%3D%5Csqrt%7B%5Cfrac%7BM_%7BO_2%7D%7D%7BM_%7BX%7D%7D%7D)
We are given:
Volume of unknown gas (X) = 1.0 L
Volume of oxygen gas = 1.0 L
Time taken by unknown gas (X) = 105 seconds
Time taken by oxygen gas = 31 seconds
Molar mass of oxygen gas = 32 g/mol
Molar mass of unknown gas (X) = ? g/mol
Putting values in above equation, we get:
![\left(\frac{\frac{1.0}{105}}{\frac{1.0}{31}}\right)=\sqrt{\frac{32}{M_X}}\\\\M_X=367.12g/mol](https://tex.z-dn.net/?f=%5Cleft%28%5Cfrac%7B%5Cfrac%7B1.0%7D%7B105%7D%7D%7B%5Cfrac%7B1.0%7D%7B31%7D%7D%5Cright%29%3D%5Csqrt%7B%5Cfrac%7B32%7D%7BM_X%7D%7D%5C%5C%5C%5CM_X%3D367.12g%2Fmol)
Hence, the molar mass of unknown gas is 367.12 g/mol