Answer:
Dissolve 47.025 grams of sucrose in enough water to produce 250 ml of solution.
Explanation:
You need to prepare 250. mL of a 0.550 M aqueous solution of sucrose, C12H22O11 (aq),
which is used frequently in biological experiments.
Based on your answer above, what is the value of x?
Solution:
A 0.550 M aqueous solution contains 0.550 mole of sucrose per liter of solution.
250 ml = 0.250 liter
Thus a 0.250 liter of a 0.550 M aqueous solution of sucrose contains 0.250 * 0.550 = 0.1375 mole of sucrose
Recall:
Mass = number of moles * mass of 1 mole
mass of 1 mole of C12H22O11
= 12(12) + 1(22) + 16(11)
= 144 + 22 + 176
= 342 grams
Mass = 0.1375 * 342 = 47.025 grams of sucrose.
Dissolve 47.025 grams of sucrose in enough water to produce 250 ml of solution.
Answer: due to insufficient exocytosis in the type II alveolar cells
the skin has cells like that
Answer:
P. aeruginosa
Explanation:
<em>P. aeruginosa</em> is a gram-negative bacteria that belongs to the family Pseudomonadaceae.
From the given question the following points lead us to conclude that the colony that will be growing would be of P. aeruginosa :
1. Flat spreading colonies with a metallic sheen on SBA - <em>P. aeruginosa</em> is known to produce smooth colonies with flat edges.
2. Fluorescent green color in the media with clear colonies on cetrimide agar - <em>P. aeruginosa</em> is known to produce pyoverdin which is a fluorescent pigment under low iron conditions.
3. Medium clear colonies that have a "fruity or grape-like odor" on MacConkey Agar - <em>P. aeruginosa</em> has a sweet fruity odor which is its characteristic odor because of the production of trimethylamine.
Thus, from all these characteristics one can conclude that the organism in the culture is <em>P. aeruginosa. </em>
