How does the process of cellular respiration create energy?

Chemistry

2 answers:

Pavlova-9 [17]3 years ago

7 0

Answer:

well, through the process of cellular respiration, the energy in food is converted into energy that can be used by the body's cells. During cellular respiration, glucose and oxygen are converted into carbon dioxide and water, and the energy is transferred to ATP

BigorU [14]3 years ago

6 0

Answer:

Oxygen is mixed with certain chemicals in a plant to make Carbon Dioxide and then the plant uses that to eat. So even if humans disappeared plants could still survive.

Explanation:

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An atom has 10 total electrons. How would you describe the reactivity of

wlad13 [49]

Answer: d

explanation:

5 0

3 years ago

The observed rotation of 2.0 gg of a compound in 50 mLmL of solution in a polarimeter tube 25-cmcm long is +47.2 ∘∘ . What is th

Anastaziya [24]

Answer:

[∝] = +472

Explanation:

Specific rotation in a solution is defined as:

[∝] = ∝ / c×l

Where:

[∝] is specific rotation, ∝ is observed rotation (In degrees), c is concentration in g/mL and l is path length (In dm).

∝: +47.2°

c: 2.0g / 50mL = 0.04g/mL

l: 25cm × (1dm /10cm) = 2.5dm

Replacing:

[∝] = +47.2° / 0.04g/mL×2.5dm = <em>+472</em>

I hope it helps!

4 0

3 years ago

If you start with 227.8 grams of iron and 128 grams of oxygen to produce iron oxide, what is the limiting reagent?

agasfer [191]

The balanced chemical equation between iron and oxygen to produce iron (III) oxide is,

$4Fe(s) + 3O_{2}(g) ---> 2Fe_{2}O_{3}(s)$

Mass of Fe = 227.8 g

Moles of Fe = $227.8 g Fe * \frac{1 mol Fe}{55.85 g Fe} = 4.079 mol Fe$

Mass of oxygen = 128 g

Moles of $O_{2} = 128 g O_{2}*\frac{1 mol O_{2}}{32 g O_{2}}= 4mol O_{2}$

Calculating the limiting reactant: The reactant that produces the least amount of product will be the limiting reactant.

Mass of iron (III) oxide produced from Iron = $4.079 mol Fe * \frac{2 mol Fe_{2}O_{3}}{4 mol Fe} *\frac{159.69 g Fe_{2}O_{3}}{1 mol Fe_{2}O_{3}} = 325.7 g Fe_{2}O_{3}$

Mass of iron (III) oxide produced from oxygen= $4 mol O_{2}*\frac{2 molFe_{2}O_{3}}{3 mol O_{2}}*\frac{159.69 g Fe_{2}O_{3}}{1 mol Fe_{2}O_{3}} = 425.84 g Fe_{2}O_{3}$