Answer:
D im pretty sure cuz i prob had this test already
Explanation:
Answer:
A.
Carbon dioxide
Explanation:
In a tissue that metabolizes glucose via the pentose phosphate pathway, C-1 of glucose would be expected to end up principally in Carbon dioxide
Answer:
5 mL
Explanation:
Given data:
mass of ring = 107 g
volume of water = 10 mL
increase in volume = 15 mL
How much water displace = ?
Solution:
V (ring) = V (water + ring) - V (water)
V (ring) = 15 mL - 10 mL
V (ring) = 5 mL
when the ring is put into cylinder, volume is increased by 15 mL. The volume of water was 10 mL so water is displaced by 5 mL and the volume 5mL is the voulme of ring.
Since you have not included the chemical reaction I will explain you in detail.
1) To determine the limiting agent you need two things:
- the balanced chemical equation
- the amount of every reactant involved as per the chemical equation
2) The work is:
- state the mole ratios of all the reactants: these are the ratios of the coefficientes of the reactans in the balanced chemical equation.
- determine the number of moles of each reactant with this formula:
number of moles = (mass in grams) / (molar mass)
- set the proportion with the two ratios (theoretical moles and actual moles)
- compare which reactant is below than the stated by the theoretical ratio.
3) Example: determine the limiting agent in this reaction if there are 100 grams of each reactant:
i) Chemical equation: H₂ + O₂ → H₂O
ii) Balanced chemical equation: 2H₂ + O₂ → 2H₂O
iii) Theoretical mole ration of the reactants: 2 moles H₂ : 1 mol O₂
iv) Covert 100 g of H₂ into number of moles
n = 100g / 2g/mol = 50 mol of H₂
v) Convert 100 g of O₂ to moles:
n = 100 g / 32 g/mol = 3.125 mol
vi) Actual ratio: 50 mol H₂ / 3.125 mol O₂
vii) Compare the two ratios:
2 mol H₂ / 1 mol O ₂ < 50 mol H₂ / 3.125 mol O₂
Conclusion: the actual ratio of H₂ to O₂ is greater than the theoretical ratio, meaning that the H₂ is in excess respect to the O₂. And that means that O₂ will be consumed completely while some H₂ will remain without react.
Therefore, the O₂ is the limiting reactant in this example.
