What are the limitations of seed banks?

1. The reaction below describes cellular respiration. Which compounds are the reactants and which are the

swat32

Answer:

C₆H₁₂O₆ and O₂ are reactant.

CO₂ and H₂O are products.

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O + ATP

Explanation:

There are two types of respiration:

1. Aerobic respiration

2. Anaerobic respiration

Aerobic respiration

It is the breakdown of glucose molecule in the presence of oxygen to yield large amount of energy. Water and carbon dioxide are also produced as a byproduct.

Glucose + oxygen → carbon dioxide + water + 38ATP

Anaerobic Respiration

It is the breakdown of glucose molecule in the absence of oxygen and produce small amount of energy. Alcohol or lactic acid and carbon dioxide are also produced as byproducts.

Glucose→ lactic acid/alcohol + 2ATP + carbon dioxide

This process use respiratory electron transport chain as electron acceptor instead of oxygen. It is mostly occur in prokaryotes. Its main advantage is that it produce energy (ATP) very quickly as compared to aerobic respiration.

Steps involve in anaerobic respiration are:

Glycolysis

Glycolysis is the first step of both aerobic and anaerobic respiration. It involve the breakdown of one glucose molecule into pyruvate and 2ATP.

Fermentation

The second step of anaerobic respiration is fermentation. It involve the fermentation of pyruvate into lactic acid or alcohol depending upon the organism in which it is taking place. There is no ATP produced, however carbon dioxide is released in this step.

8 0

3 years ago

Please help with #2 and #3

Vaselesa [24]

Answer:

2. $V_2=17L$

3. $V=82.9L$

Explanation:

Hello there!

2. In this case, we can evidence the problem by which volume and temperature are involved, so the Charles' law is applied to:

$\frac{V_2}{T_2}=\frac{V_1}{T_1}$

Thus, considering the temperatures in kelvins and solving for the final volume, V2, we obtain:

$V_2=\frac{V_1T_2}{T_1}$

Therefore, we plug in the given data to obtain:

$V_2=\frac{18.2L(22+273)K}{(45+273)K} \\\\V_2=17L$

3. In this case, it is possible to realize that the 3.7 moles of neon gas are at 273 K and 1 atm according to the STP conditions; in such a way, considering the ideal gas law (PV=nRT), we can solve for the volume as shown below:

$V=\frac{nRT}{P}$