Answer:
0.825 M
Explanation:
The osmotic pressure is a colligative property, that can be calculated using the following expression.
π = M × R × T
where,
π is the osmotic pressure
M is the molarity
R is the ideal gas constant
T is the absolute temperature (24°C + 273 = 297 K)
M = π / R × T = 20.1 atm / (0.08206 atm.L/mol.K) × 297 K = 0.825 M
Given what we know, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.
<h3>Why does it take this much energy to boil the water?</h3>
We arrive at this number by taking into account the energy needed to boil 1g of water to its vaporization point. This results in the use of 2260 J of heat energy. We then take this number and multiply it by the total grams of water being heated, in this case, 5.05g, which gives us our answer of 11.4 kJ of energy required.
Therefore, we can confirm that the amount of heat energy that would be required in order to boil 5.05g of water is that of 11.4kJ of heat.
To learn more about the behavior of water visit:
brainly.com/question/1416592?referrer=searchResults
The answer is C. Hydrogen Bond
The balanced reaction equation for the combustion of butane is as follows;
C₄H₁₀ + 13/2O₂ ---> 4CO₂ + 5H₂O
the limiting reactant in this reaction is C₄H₁₀ This means that all the butane moles are consumed and amount of product formed depends on the amount of C₄H₁₀ used up.
stoichiometry of C₄H₁₀ to H₂O is 1:5
mass of butane used - 6.97 g
number of moles - 6.97 g / 58 g/mol = 0.12 mol
then the number of water moles produced - 0.12 mol x 5 = 0.6 mol
Therefore mass of water produced - 0.6 mol x 18 g/mol = 10.8 g
