Butylated hydroxytoluene (BHT) is used as an antioxidant in processed foods. (It prevents fats and oils from becoming rancid.) A
1 answer:
Answer:
Molecular mass of BHT is 204,2 g/mol
Explanation:
To answer this question you need to know that Freezing-point depression (A colligative property) is the decrease of the freezing point of a solvent on the addition of a non-volatile solute. The formula is:
ΔT = Kf m
Where ΔT is the freezing point depression (5,5°C - 4,880°C = 0,62°C
Kf that is freezing point molar constant of the solvent (5,065°C/m)
And m that is molality (moles of solute/kg of solvent). Replacing:
Knowing you have 100,0g of solvent ≡ 0,1000 kg:
=<em> 0,01224 moles of solute </em> <em>≡ BHT</em>
As grams of BHT are 2,500, molecular mass of BHT is:
<em>204,2 g/mol</em>
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I hope it helps
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Answer: 0.596
Explanation:
For this problem, we want to find . To do so, we will need to use the ICE chart. The I in ICE is initial quantity. In this case, it is the initial pressure. Pressure is in atm. The C in ICE is change in each quantity. The E is equilibrium.
H₂(g) + I₂(g) ⇄ 2HI(g)
I 3.12 5.52 0
C -0.869 -0.869 +1.738
E 2.251 2.251 1.738
<u>For the steps below, refer to the ICE chart above.</u>
1. Since we were given the initial of H₂, I₂ and equilibrium of 2HI, we can fill those into the chart.
2. Since we were not given the initial for 2HI, we will put 0 in their place.
3. For the change, we need to add pressure to the products to make the reaction reach equilibrium. We would add on the products and subtract from the reactants to equalize the reaction. Since we don't know how much the change in, we can use variable x. We know that the equilibrium of 2HI is 1.738, we know the change is 0.869 because 1.738/2=0.869. Since there are 2 moles of HI, we must divide the equilibrium by 2 to find x, so that we can fill that into the reactants side.
4. With the equilibrium values, we can find the equilibrium pressure. It is products over reactants. We use the values of E.
The [HI]² comes from 2HI. We more the moles to the exponent when we are calculating the equilibrium pressure.
We typically don't put units because K is unitless, but know that the units is atm throughout the entire problem.