D, carbon dioxide and water. Methane and oxygen PRODUCE carbon dioxide and water in their reaction.
To calculate the molarity you only need to know the number of moles in the solution and the volume of that solution. This exercise gives both and with that you divide moles by volume(usually in liters).
500 ml equals 0,5 L
molarity= number of moles/ volume
molarity=0,75 x 0,5
= 0,375 mol/L
Answer:
Electron transport is the process by NADH + H+ and FADH2 are converted to NAD+ and FAD, donating electrons and hydrogen ions to oxygen
Explanation:
The limiting reactant is chlorine (Cl2).
<u>Explanation</u>:
Limiting reactant is the amount of product formed which gets limited by the reagent without continuing it.
2 Al + 3 Cl2 ==> 2 AlCl3 represents the balanced equation.
Number of moles Al present = 34 g Al x 1 mole Al / 26.98 g
= 1.260 g moles of Al
Number of moles Cl2 present = 39 g Cl2 x 1 mole Cl2 / 35.45 g
= 1.10 g moles of Cl2
Dividing each reactant by it's coefficient in the balanced equation obtains:
1.260 moles Al / 2 = 0.63 g moles of Al
1.11 moles Cl2 / 3 = 0.36 g moles of Cl2
The reactant which produces a lesser amount of product is called as limiting reactant.
Here the Limiting reactant is Cl2.
The mass of water that contains 2.5×10²⁴ atoms of Hydrogen is 74.79 g
<h3>Avogadro's hypothesis </h3>
From Avogadro's hypothesis,
6.02×10²³ atoms = 2 g of H
Therefore,
2.5×10²⁴ atoms = (2.5×10²⁴ × 2) / 6.02×10²³
2.5×10²⁴ atoms = 8.31 g of H
<h3>How to determine the mass of water </h3>
- 1 mole of water H₂O = (2×1) + 16 = 18 g
- Mass of H in 1 mole of water = 2 g
2 g of H is present in 18 g of water.
Therefore,
8.31 g of H will be present in = (8.31 × 18) / 2 = 74.79 g of water.
Thus, 2.5×10²⁴ atoms of Hydrogen is present in 74.79 g of water.
Learn more about Avogadro's number:
brainly.com/question/26141731
