All categories

3 years ago

What are the three end products of the oxidation-reduction reactions involved in metabolism?

1 answer:

7 0

Answer: The three end products of the oxidation-reduction reactions involved in metabolism are WATER, CARBON DIOXIDE and ATP.

Metabolism refers to the sum of all chemical reactions occurring in the body. It includes both catabolism and anabolism. Every metabolic reaction is characterized by oxidation-reduction reactions. The metabolic pathways include respiration, photosynthesis, etc. In all these reactions water is produced. In respiration CO2 is also a product. In general. in every metabolic reaction either an organic molecule is broken or synthesized producing CO2, water and energy.

You might be interested in

Anna studies the color of her bedroom wall using different lights. In sunlight, she sees that the wall is white. Anna darkens th

Marizza181 [45]

Answer: C white

Explanation:

4 0

3 years ago

Read 2 more answers

A submarine changes its mass by adding or removing seawater from tanks inside the sub. Explain how this can enable the sub to di

bogdanovich [222]

Less water = less weight to make it rise
More water = more weight to make it dive

3 0

3 years ago

What would happen if the Earth was tilted more than its current tilt of 23.5°? <br><br> HELP ASAP!!

Oduvanchick [21]

Answer:

Wouldn't the Earth's atmosphere be moving too fast that it eventually breaks out?

Explanation:

Do NOT trust me.

7 0

3 years ago

a sample of compound determined to contain 1.71 g C and 0.287 g H. The corresponding empirical formula is

fiasKO [112]

Answer: The empirical formula is $CH_2$ .

Explanation:

Mass of C = 1.71 g

Mass of H = 0.287 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{1.71g}{12g/mole}=0.142moles$

Moles of H = $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.287g}{1g/mole}=0.287moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.142}{0.142}=1$

For H = $\frac{0.287}{0.142}=2$

The ratio of C: H = 1: 2

Hence the empirical formula is $CH_2$ .

4 0

3 years ago

Gallium has two naturally occurring isotopes: 69ga with a mass of 68.9256 amu and a natural abundance of 60.11% and 71ga. use th

Nataly [62]

There are two naturally occurring isotopes of gallium: mass of Ga-69 isotope is 68.9256 amu and its percentage abundance is 60.11%, let the mass of other isotope that is Ga-71 be X, the percentage abundance can be calculated as:

%Ga-71=100-60.11=39.89%

Atomic mass of an element is calculated by taking sum of atomic masses of its isotopes multiplied by their percentage abundance.

Thus, in this case:

Atomic mass= m(Ga-69)×%(Ga-69)+X×%(Ga-71)

From the periodic table, atomic mass of Ga is 69.723 amu.

Putting the values,

$69.723 amu=(68.9256 amu)(\frac{60.11}{100})+X(\frac{39.89}{100})$