Answer:
The answer to your question is 0.67 moles of Glucose
Explanation:
Data
120 g of Glucose
convert to moles
Process
1.- Calculate the molar mass of Glucose
C₆H₁₂O₆ = (12 x 6) + (1 x 12) + (16 x 6)
= 72 + 12 + 96
= 180 g
2.- Remember that the molar mass of each molecule equals 1 mol. Use the rule of three to find the answer.
180 g ------------------- 1 mol
120 g ------------------ x
x = (120 x 1)/180
x = 120/180
x = 0.67 moles
3.- Conclusion
120 g of Glucose are equal to 0.67 moles.
1. “what forms of energy conversions occur during the process of photosynthesis? (How does energy transform?) 2. What is missing from the food web but is essential to maintain equilibrium? A. Soil B.water C. Decomposers D. Oxygen
Answer:
hydrogen gas
Explanation:
Fuel cell vehicles use hydrogen gas to power an electric motor. Unlike conventional vehicles which run on gasoline or diesel, fuel cell cars and trucks combine hydrogen and oxygen to produce electricity, which runs a motor.
Answer:
1) The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
2) The amount (in grams) of excess reactant H₂ = 4.39 g.
Explanation:
- Firstly, we should write the balanced equation of the reaction:
<em>N₂ + 3H₂ → 2NH₃.</em>
<em>1) To determine the limiting reactant of the reaction:</em>
- From the stichiometry of the balanced equation, 1.0 mole of N₂ reacts with 3.0 moles of H₂ to produce 2.0 moles of NH₃.
- This means that <em>N₂ reacts with H₂ with a ratio of (1:3).</em>
- We need to calculate the no. of moles (n) of N₂ (5.23 g) and H₂ (5.52 g) using the relation:<em> n = mass / molar mass.</em>
The no. of moles of N₂ in (5.23 g) = mass / molar mass = (5.23 g) / (28.00 g/mol) = 0.1868 mol.
The no. of moles of H₂ (5.52 g) = mass / molar mass = (5.52 g) / (2.015 g/mol) = 2.74 mol.
- From the stichiometry, N₂ reacts with H₂ with a ratio of (1:3).
The ratio of the reactants of N₂ (5.23 g, 0.1868 mol) to H₂ (5.52 g, 2.74 mol) is (1:14.67).
∴ The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
<em>2) To determine the amount (in grams) of excess reactant of the reaction:</em>
- As showed in the part 1, The limiting reactant is N₂ because it is present with the lower no. of moles than H₂.
- Also, 0.1868 mol of N₂ react completely with 0.5604 mol of H₂ and the remaining of H₂ is in excess.
- The no. of moles are in excess of H₂ = 2.74 mol - 0.5604 mol (reacted with N₂) = 2.1796 mol.
- ∴ The amount (in grams) of excess reactant H₂ = n (excess moles) x molar mass = (2.1796 mol)((2.015 g/mol) = 4.39 g.
Answer: The conversion to malate
Explanation:
Pyruvate is the process which produced in glycoysis which has multiple fates and it can give rises to acetyl co-enzyme and undergo the aerobic oxidation in the critic acid cycle. It can be used to produces glucose but it never produced the malate. In prokaryotes it can be processes in the anaerobic respiration to produced the ethanol, as end product.
