It is limited by the depth divers and equipment can go.
<span>K D = 1.0g/ 28 ml 0.0357
-------------- = ---------- = 1.65
1.0 g/46 ml 0.0217
is the distribution coefficient of caffeine in this solvent system</span>
Answer:
1) When 69.9 g heptane is burned it releases 5.6 mol water.
2) C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.
Explanation:
- Firstly, we should balance the equation of heptane combustion.
- The balanced equation is: <em>C₇H₁₆ + 11O₂ → 7CO₂ + 8H₂O.</em>
This means that every 1.0 mole of complete combustion of heptane will release 8 moles of H₂O.
- We need to calculate the no. of moles in 69.9 g of heptane that is burned using the relation: <em>n = mass/molar mass.</em>
n of 69.9 g of heptane = mass/molar mass = (69.9 g)/(100.21 g/mol) = 0.697 mol ≅ 0.7 mol.
<em><u>Using cross multiplication:</u></em>
1.0 mol of heptane releases → 8 moles of water.
0.7 mol of heptane releases → ??? moles of water.
<em>∴ The no. of moles of water that will be released from burning (69.9 g) of water</em> = (0.7 mol)(8.0 mol)/(1.0 mol) = <em>5.6 mol.</em>
<em>∴ When 69.9 g heptane is burned it releases </em><em>5.6</em><em> mol water. </em>
<em />
Answer:
Up to 80 - 120 days.
Explanation:
The flower will probably stay six to twelve days or so.
I'm not sure what's your hypothesis going to be, but I'll give you an example.
" <em>If</em> I __________, <em>then</em> the sunflower will grow up to 80 to 120 days."
In the blank space, you can write what you're going to do to your sunflower during the experiment.
Please correct me if I'm wrong.
