Answer: 120g/mol
Explanation:
The first step we are to take is to calculate the freezing point depression of the solution.
ΔT(f) = freezing point of pure solvent - freezing point of solution
ΔT(f) = 5.48 - 3.77
ΔT(f) = 1.71°C
Next we are to calculate the molal concentration of the solution using freezing point depression
ΔT(f) = K(f) * m
m = ΔT(f)/K(f)
m = 1.71/5.12
m = 0.333 molal
Now, we calculate the molecular weight of the unknown...
m = 0.333 mol = 0.333 mol X per kg of benzene
moles of X = 0.333 mol of X per kg of benzene * 0.5kg of benzene
moles of X = 0.1665
molecular weight of X = 20g of X/0.1665
molecular weight of X = 120/mol
Answer:
207.89g
Explanation:
The formula of the compound is:
Fe₂S₃
Find the molar mass of the compound;
Atomic mass of Fe = 55.845g/mol
S = 32.065g/mol
Now;
Molar mass of Fe₂S₃ = 2 (55.845) + 3 (32.065)
= 207.89g
<span>The molar mass of the compound is 36.355 g/mol. This is calculated by knowing that 1 mol of gas fills 22.4 L of volume, so 1.623 g/L = X g/mol * 1/22.4 mol/L -> 1.623 g/L * 22.4 L/mol = X g/mol -> 36.355 g/mol = X g/mol</span>
Answer:
The least substituted product (anti-Markovnikov)
Explanation:
The ROOR is used in the addition reaction of HBr to an organic substance (an alkene for example).
In normal conditions (with no ROOR) the adition of the halogen will be performed in the most substituted C (following the rule of Markovnikov that says that the stability increases with the more substituted is the C).
But in presence of ROOR, the reaction takes other mechanism (free radicals), and the product in this case is the one with the Br added in the least substituted C.
Answer:
The answer to your question is [H₃O⁺] = 0.025 [OH⁻] = 3.98 x 10⁻¹³
Explanation:
Data
[H⁺] = ?
[OH⁻] = ?
pH = 1.6
Process
Use the pH formula to calculate the [H₃O⁺], then calculate the pOH and with this value, calculate the [OH⁻].
pH formula
pH = -log[H₃O⁺]
-Substitution
1.6 = -log[H₃O⁺]
-Simplification
[H₃O⁺] = antilog (-1,6)
-Result
[H₃O⁺] = 0.025
-Calculate the pOH
pOH = 14 - pH
-Substitution
pOH = 14 - 1.6
-Result
pOH = 12.4
-Calculate the [OH⁻]
12.4 = -log[OH⁻]
-Simplification
[OH⁻] = antilog(-12.4)
-Result
[OH⁻] = 3.98 x 10⁻¹³
