Do we get multiple choice
B.
about the same as for humans
Explanation:
This is why most bacteria, during research in labs, are incubated at 37 degrees centigrade, about the same as human body temperature. In addition, harmful and beneficial bacteria thrive in the human body due to the favorable temperatures for growth and reproduction. To try and fight an infection, the body also tries to raise the body temperatures above the optimal for the bacteria growth (the reason one has a fever in case of infection).
The answer is most definitely “A”
Answer:
C₅H₁₀O₅
Explanation:
1. Calculate the mass of each element in 2.78 mg of X.
(a) Mass of C

(b) Mass of H

(c) Mass of O
Mass of O = 3.5 - 1.400 - 0.2349 = 1.87 g
2. Calculate the moles of each element

3. Calculate the molar ratios
Divide all moles by the smallest number of moles.

4. Round the ratios to the nearest integer
C:H:O = 1:2:1
5. Write the empirical formula
The empirical formula is CH₂O.
6. Calculate the molecular formula.
EF Mass = (12.01 + 2.016 + 16.00) u = 30.03 u
The molecular formula is an integral multiple of the empirical formula.
MF = (EF)ₙ

MF = (CH₂O)₅ = C₅H₁₀O₅
The molecular formula of X is C₅H₁₀O₅.
<u>Answer:</u> The pH of the buffer is 4.61
<u>Explanation:</u>
To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:
![pH=pK_a+\log(\frac{[\text{conjuagate base}]}{[\text{acid}]})](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%28%5Cfrac%7B%5B%5Ctext%7Bconjuagate%20base%7D%5D%7D%7B%5B%5Ctext%7Bacid%7D%5D%7D%29)
We are given:
= negative logarithm of acid dissociation constant of weak acid = 4.70
= moles of conjugate base = 3.25 moles
= Moles of acid = 4.00 moles
pH = ?
Putting values in above equation, we get:

Hence, the pH of the buffer is 4.61