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Advocard [28]
3 years ago
6

Assume you are working for a chemical company and are responsible for growing a yeast culture that produces ethanol. The yeasts

are growing well on the maltose medium but are not producing alcohol. What is the most likely explanation?
Chemistry
1 answer:
romanna [79]3 years ago
6 0

Answer:

Respiration might occur in the presence and absence of oxygen known as the aerobic respiration and anaerobic respiration. The ATP molecules are produced more in number in aerobic respiration.

The yeast has the ability to undergo the process of anaerobic respiration and its end product are alcohol, carbon dioxide and 2 moles of ATP. The yeast is grown on maltose medium but unable to produce alcohol because of the presence of oxygen in the medium. The oxygen might acts as poison or inhibit the process of anaerobic respiration.

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A 44.8 g rhodium sample contains how many rhodium atoms
Tresset [83]

Answer:

2.63*10^23

Explanation:

1 mol rhodium = 102.91

44.8g/1 mol * 1 mol/ 102.91mol * 6.022*10^23/1 mol =

2.63*10^23

6 0
3 years ago
A compound is found to be 30.45% n and 69.55 % o by mass. if 1.63 g of this compound occupy 389 ml at 0.00°c and 775 mm hg, what
Charra [1.4K]
1) mass composition

N: 30.45%
O: 69.55%
   -----------
   100.00%

2) molar composition

Divide each element by its atomic mass

N: 30.45 / 14.00 = 2.175 mol

O: 69.55 / 16.00 = 4.346875

4) Find the smallest molar proportion

Divide both by the smaller number

N: 2.175 / 2.175 = 1

O: 4.346875 / 2.175 = 1.999 = 2

5) Empirical formula: NO2

6) mass of the empirical formula

14.00 + 2 * 16.00 = 46.00 g

7) Find the number of moles of the gas using the equation pV = nRT

=> n = pV / RT = (775/760) atm * 0.389 l / (0.0821 atm*l /K*mol * 273.15K)

=> n = 0.01769 moles

8) Find molar mass

molar mass = mass in grams / number of moles = 1.63 g / 0.01769 mol = 92.14 g / mol

9) Find how many times the mass of the empirical formula is contained in the molar mass

92.14 / 46.00 = 2.00

10) Multiply the subscripts of the empirical formula by the number found in the previous step

=> N2O4

Answer: N2O4
3 0
3 years ago
Chemistry. Will mark Brainly.
kogti [31]

Answer:

c is right

Explanation:

8 0
3 years ago
Draw the most stable resonance structure for the intermediate in the electrophilic aromatic bromination of aniline, anisole, and
ASHA 777 [7]

Answer:

Here's what I get

Explanation:

(a) Intermediates

The three structures below represent one contributor to the resonance-stabilized intermediate, in which the lone pair electrons on the heteroatom are participating (the + charge on the heteroatoms do not show up very well).

(b) Relative Stabilities

The relative stabilities decrease in the order shown.

N is more basic than O, so NH₂ is the best electron donating group (EDG) and will best stabilize the positive charge in the ring. However, the lone pair electrons on the N in acetanilide are also involved in resonance with the carbonyl group, so they are not as available for stabilization of the ring.

(c) Relative reactivities

The relative reactivities would be

C₆H₅-NH₂ >  C₆H₅-OCH₃ > C₆H₅-NHCOCH₃

4 0
3 years ago
Based on the graph below, if you chose an alien at random on day 20, what form would it most likely be in?
ArbitrLikvidat [17]

Answer:

A. Thin

Explanation:

i took the test a while back yw <3

4 0
3 years ago
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