Answer:
carbon dioxide , water , 38ATP
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.
They all have endoskeletons that are formed of bone or cartilage. Hope this helped! (:
Answer: its either argon or phosphorus is this a constructed response or a multiple choice
Explanation:
When solutions of sodium sulfide and copper(ii) sulfate are mixed, a precipitate of copper(ii) sulfide is formed. The net ionic equation for this reaction is Cu⁺² (aq) + S⁻² (aq) → CuS (s).
<h3>What is Balanced Chemical Equation ?</h3>
The balanced chemical equation is the equation in which the number of atoms on the reactant side is equal to the number of atoms on the product side in an equation.
Now first write the balanced chemical equation
CuSO₄ (aq) + Na₂S (aq) → CuS(s) + Na₂SO₄ (aq)
Now write the net ionic equation
Cu⁺² (aq) + SO₄⁻² (aq) + 2Na⁺ (aq) + S⁻² (aq) → CuS (s) + 2Na⁺ + SO₄⁻² (aq)
So the net ionic equation is
Cu⁺² (aq) + S⁻² (aq) → CuS (s)
Thus from the above conclusion we can say that When solutions of sodium sulfide and copper(ii) sulfate are mixed, a precipitate of copper(ii) sulfide is formed. The net ionic equation for this reaction is Cu⁺² (aq) + S⁻² (aq) → CuS (s).
Learn more about the Balanced Chemical Equation here: brainly.com/question/26694427
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Answer:
C8H17N
Explanation:
Mass of the unknown compound = 5.024 mg
Mass of CO2 = 13.90 mg
Mass of H2O = 6.048 mg
Next, we shall determine the mass of carbon, hydrogen and nitrogen present in the compound. This is illustrated below:
For carbon, C:
Molar mass of CO2 = 12 + (2x16) = 44g/mol
Mass of C = 12/44 x 13.90 = 3.791 mg
For hydrogen, H:
Molar mass of H2O = (2x1) + 16 = 18g/mol
Mass of H = 2/18 x 6.048 = 0.672 mg
For nitrogen, N:
Mass N = mass of unknown – (mass of C + mass of H)
Mass of N = 5.024 – (3.791 + 0.672)
Mass of N = 0.561 mg
Now, we can obtain the empirical formula for the compound as follow:
C = 3.791 mg
H = 0.672 mg
N = 0.561 mg
Divide each by their molar mass
C = 3.791 / 12 = 0.316
H = 0.672 / 1 = 0.672
N = 0.561 / 14 = 0.040
Divide by the smallest
C = 0.316 / 0.04 = 8
H = 0.672 / 0.04 = 17
N = 0.040 / 0.04 = 1
Therefore, the empirical formula for the compound is C8H17N
