Answer:
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Explanation:
The lowering of the freezing point of a solvent is a colligative property ruled by the formula:
Where:
- ΔTf is the lowering of the freezing point
- Kf is the molal freezing constant of the solvent: 1.86 °C/m
- m is the molality of the solution
- i is the van't Hoff factor: the number of particles (ions) per unit of ionic compound.
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<u>a) molality, m</u>
- m = number of moles of solute/ kg of solvent
- number of moles of CaI₂ = mass in grams/ molar mass
- number of moles of CaI₂ = 25.00g / 293.887 g/mol = 0.0850667mol
- m = 0.0850667mol/1.25 kg = 0.068053m
<u>b) i</u>
- Each unit of CaI₂, ideally, dissociates into 1 Ca⁺ ion and 2 I⁻ ions. Thus, i = 1 + 2 = 3
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<u>c) Freezing point lowering</u>
- ΔTf = 1.86 °C/m × 0.068053m × 3 = 0.3797ºC ≈ 0.380ºC
<h2>I have problems to upload the full answer in here, so I attach a pdf file with the whole answer.</h2>
Answer:
Mass = 0.139 kg
Explanation:
There is a formula in terms of force, acceleration and mass which is:
Force = mass × acceleration
Put the values into the formula.
5 = m × 36
m = 5 ÷ 36
<u>m = 0.139 kg</u>
Sodium, Atomic mass: 22.989769 g
You can see in a periodic table
Answer : The molecular formula of the compound is, 
Explanation :
Molecular formula : It is the representation of substance by the symbols and it denotes the number of atoms of each element present in the compound.
Now count the number of carbon, hydrogen, nitrogen and oxygen atoms present in the given compound.
As we see that in the given compound, there are 8 atoms of carbon element, 3 atoms of oxygen element, 1 atom of nitrogen element, 9 atoms of hydrogen element.
Thus, the molecular formula of the compound will be 
Answer:
Option (1) Br– is the catalyst, and the reaction follows a faster pathway with Br– than without
Explanation:
Let us consider the equation below:
Step 1:
H2O2(aq) + Br–(aq) → H2O(l) + BrO–(aq)
Step 2:
BrO–(aq) + H2O2(aq) → H2O(l) + O2(g) + Br–(aq)
From the above equation, we can see that Br– is unchanged.
This implies that Br– is the catalyst as catalyst does not take part in a chemical reaction but they create an alternate pathway to lower the activation energy in order for the reaction to proceed at a much faster rate to arrive at the products.