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3 years ago

What are the products when anaerobic respiration occurs in yeast cells?

Chemistry

2 answers:

DochEvi [55]3 years ago

7 0

<span>carbon dioxide, water, and ATP

</span>

kicyunya [14]3 years ago

5 0

The correct answer is the second option. Anaerobic respiration is a type of respiration which involves the use of electron acceptors other than oxygen. This is respiration without using oxygen. An example is alcohol fermentation where the the reactants are glucose and enzymes forming to products which are CO2, ethanol and energy.

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Flame tests are used to identify

sergij07 [2.7K]

Answer:

Explanation: The flame test is used to visually determine the identity of an unknown metal or metalloid ion based on the characteristic color.

Hope this helps

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3 years ago

Read 2 more answers

Solid to Gas

miskamm [114]

Answer:

c. c. c c c c c cçcc. ccccccccc

Explanation:

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2 years ago

What is the valence of K2O2

raketka [301]

14 since K has 1 valence but there’s two so 2 valence for k and oxygen has 6 but there’s two so 12

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3 years ago

ubstance A undergoes a first order reaction A®B with a half-life, t½, of 20 min at 25 °C. If the initial concentration of A in a

Stells [14]

Answer : The concentration of A after 80 min is, 0.100 M

Explanation :

Half-life = 20 min

First we have to calculate the rate constant, we use the formula :

$k=\frac{0.693}{t_{1/2}}$

$k=\frac{0.693}{20\text{ min}}$

$k=3.465\times 10^{-2}\text{ min}^{-1}$

Expression for rate law for first order kinetics is given by:

$t=\frac{2.303}{k}\log\frac{a}{a-x}$

where,

k = rate constant = $3.465\times 10^{-2}\text{ min}^{-1}$

t = time passed by the sample = 80 min

a = initial amount of the reactant = 1.6 M

a - x = amount left after decay process = ?

Now put all the given values in above equation, we get

$80=\frac{2.303}{3.465\times 10^{-2}}\log\frac{1.6}{a-x}$

$a-x=0.100M$

Therefore, the concentration of A after 80 min is, 0.100 M

3 0

2 years ago

Air, with an initial absolute humidity of 0.016 kg water per kg dry air, is to be dehumidified for a drying process by an air co

amid [387]

Answer:

(a) The proportion of dry air bypassing the unit is 14.3%.

(b) The mass of water removed is 1.2 kg per 100 kg of dry air.

Explanation:

We can express the proportion of air that goes trough the air conditioning unit as $p_{d}$ and the proportion of air that is by-passed as $p_{bp}$ , being $p_{d}+p_{bp}=1$ .

The amount of water that goes into the drier inlet has to be 0.004 kg/kg, and can be expressed as:

$0.004 = 0.016*p_{bp}+ 0.002*p_{d}$

Replacing the first equation in the second one we have

$0.004 = 0.016*(1-p_{d})+ 0.002*p_{d}=0.016-0.016*p_{d}+0.002*p_{d}\\0.004 - 0.016 = (-0.016+0.002)*p_{d}\\-0.012 = -0.014*p_{d}\\p_{d}=\frac{-0.012}{-0.014}=0.857\\\\p_{bp}=1-p_{d}=1-0.857=0.143$

(b) Of every kg of dry air feed, 85.7% goes in to the air conditioning unit.

It takes (0.016-0.002)=0.014 kg water per kg dry air feeded.

The water removed of every 100 kg of dry air is

$100 kgDA*0.857*0.014 kgW/kgDA= 1.1998 \approx 1.2 kgW$

It can also be calculated as the difference in humiditiy between the inlet and the outlet: (0.016-0.004=0.012 kgW/kDA) and multypling by the total amount of feed (100 kgDA).

100 * 0.012 = 1.2 kgW

7 0

3 years ago