Answer:
u already know the answer
Explanation:answer
The molality of a solute is equal to the moles of solute per kg of solvent. We are given the mole fraction of I₂ in CH₂Cl₂ is <em>X</em> = 0.115. If we can an arbitrary sample of 1 mole of solution, we will have:
0.115 mol I₂
1 - 0.115 = 0.885 mol CH₂Cl₂
We need moles of solute, which we have, and must convert our moles of solvent to kg:
0.885 mol x 84.93 g/mol = 75.2 g CH₂Cl₂ x 1 kg/1000g = 0.0752 kg CH₂Cl₂
We can now calculate the molality:
m = 0.115 mol I₂/0.0752 kg CH₂Cl₂
m = 1.53 mol I₂/kg CH₂Cl₂
The molality of the iodine solution is 1.53.
Answer:
35.75 days
Explanation:
From the given information:
For first-order kinetics, the rate law can be expressed as:
Given that:
the rate degradation constant = 0.12 / day
current concentration C = 0.05 mg/L
initial concentration C₀ = 3.65 mg/L
㏑(0.01369863014) = -(0.12) t
-4.29 = -(0.12)
t = -4.29/-0.12
t = 35.75 days
Answer:
The answer is option 3.
Explanation:
When salt is added to the water, the boiling point increases because it needs to take in more energy from heat to <u>b</u><u>r</u><u>e</u><u>a</u><u>k</u><u> </u><u>d</u><u>o</u><u>w</u><u>n</u> the bonds and dissolve the salt in the water.
(Correct me if I am wrong)
The question here is solved using basic chemistry. CaCl2(aq) is an ionic compound which will have the releasing of 2 Cl⁻ ions ions in water for every molecule of CaCl2 that dissolves.
CaCl2(s) --> Ca+(aq) + 2 Cl⁻(aq)
[Cl⁻] = 0.65 mol CaCl2/1L × 2 mol Cl⁻ / 1 mol CaCl2 = 1.3 M
The answer to this question is [Cl⁻] = 1.3 M
