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3 years ago

Which of the following is a product of aerobic respiration?

Chemistry

2 answers:

sasho [114]3 years ago

3 0

Answer:

The product of aerobic respiration is Carbon dioxide.

Explanation:

The process of breaking down glucose to produce energy and waste products is called respiration. Livings beings need respiration process to generate energy so that they can survive.
The types of respiration are : Anaerobic and aerobic respiration.
Aerobic respiration takes place in presence of oxygen and produces large amount of energy.
The final product of aerobic respiration are carbon dioxide, water and 38 ATP of energy.

Anton [14]3 years ago

3 0

Answer:

The correct answer is carbon dioxide and water! I have the question on study island and got it correct! Have a wonderful day

Explanation:

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What percentage of the filtrate's water that enters bowman's capsule is reabsorbed into the blood? select one:

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2 years ago

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3 years ago

Provide a balanced molecular equation, total ionic, and net ionic equation for sodium phosphate and zinc acetate.

vivado [14]

Answer: Balanced molecular equation :

$2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)$

Total ionic equation:

$6Na^+(aq)+3PO_4^{2-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)$

The net ionic equation:

$2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)$

Explanation:

Complete ionic equation : In complete ionic equation, all the substances that are strong electrolyte are present in an aqueous state as ions.

Net ionic equation : In the net ionic equations, we are not include the spectator ions in the equations.

Spectator ions : The ions present on reactant and product side which do not participate in a reactions. The same ions present on both the sides.

When sodium phosphate and zinc acetate then it gives zinc phosphate and sodium acetate as product.

The balanced molecular equation will be,

$2Na_3PO_4(aq)+3(CH_3COO)_2Zn(aq)\rightarrow 6CH_3COONa(aq)+Zn_3(PO_4)_2(s)$

The total ionic equation in separated aqueous solution will be,

$6Na^+(aq)+2PO_4^{3-}(aq)+6CH_3COO^-(aq)+3Zn^{2+}(aq)\rightarrow 6CH_3COO^-(aq)+6Na^+(aq)+Zn_3(PO_4)_2(s)$

In this equation, and are the spectator ions.

By removing the spectator ions from the balanced ionic equation, we get the net ionic equation.

The net ionic equation will be,

$2PO_4^{3-}(aq)+3Zn^{2+}(aq)\rightarrow Zn_3(PO_4)_2(s)$

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2 years ago