The question is incomplete, the complete question is:
A certain liquid X has a normal freezing point of 
and a freezing point depression constant 
. Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.
<u>Answer:</u> The freezing point of the solution is 
<u>Explanation:</u>
Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.
The expression for the calculation of depression in freezing point is:
OR
......(1)
where,
Freezing point of pure solvent =
Freezing point of solution = 
i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)
= freezing point depression constant = 
= Given mass of solute (iron (III) chloride) = 81.1 g
= Molar mass of solute (iron (III) chloride) = 162.2 g/mol
= Mass of solvent (X) = 850. g
Putting values in equation 1, we get:
![0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC](https://tex.z-dn.net/?f=0.8-%28%5Ctext%7BFreezing%20point%20of%20solution%7D%29%3D4%5Ctimes%207.82%5Ctimes%20%5Cfrac%7B81.1%5Ctimes%201000%7D%7B162.2%5Ctimes%20850%7D%5C%5C%5C%5C%5Ctext%7BFreezing%20point%20of%20solution%7D%3D%5B0.8-18.4%5D%5EoC%5C%5C%5C%5C%5Ctext%7BFreezing%20point%20of%20solution%7D%3D-17.6%5EoC)
Hence, the freezing point of the solution is
Slow chemical change
It is a chemical change because the erosion is due to the chemical reaction between the acid and the in the rain and the calcium carbonate.
It is slow due to the concentration of acid is low.
Answer:
glucose
Explanation:
the main outputs are oxygen and glucose sugars
Answer:
30.4 g. NH3
Explanation:
This problem tells us that the hydrogen (H2) is the limiting reactant, as there is "an excess of nitrogen." Using stoichiometry (the relationship between the various species of the equation), we can see that for every 3 moles of H2 consumed, 2 moles of NH3 are produced.
But before we can use that relationship to find the number of grams of ammonia produced, we need to convert the given grams of hydrogen into moles:
5.4 g x [1 mol H2/(1.008x2 g.)] = 2.67857 mol H2 (not using significant figures yet; want to be as accurate as possible)
Now, we can use the relationship between H2 and NH3.
2.67857 mol H2 x (2 mol NH3/3 mol H2) = 1.7857 mol NH3
Now, we have the number of moles of ammonia produced, but the answer asks us for grams. Use the molar mass of ammonia to convert.
1.7857 mol NH3 x 17.034 g. NH3/mol NH3 = 30.4 g. NH3 (used a default # of 3 sig figs)
Answer:
A. ΔG° = 132.5 kJ
B. ΔG° = 13.69 kJ
C. ΔG° = -58.59 kJ
Explanation:
Let's consider the following reaction.
CaCO₃(s) → CaO(s) + CO₂(g)
We can calculate the standard enthalpy of the reaction (ΔH°) using the following expression.
ΔH° = ∑np . ΔH°f(p) - ∑nr . ΔH°f(r)
where,
n: moles
ΔH°f: standard enthalpy of formation
ΔH° = 1 mol × ΔH°f(CaO(s)) + 1 mol × ΔH°f(CO₂(g)) - 1 mol × ΔH°f(CaCO₃(s))
ΔH° = 1 mol × (-635.1 kJ/mol) + 1 mol × (-393.5 kJ/mol) - 1 mol × (-1206.9 kJ/mol)
ΔH° = 178.3 kJ
We can calculate the standard entropy of the reaction (ΔS°) using the following expression.
ΔS° = ∑np . S°p - ∑nr . S°r
where,
S: standard entropy
ΔS° = 1 mol × S°(CaO(s)) + 1 mol × S°(CO₂(g)) - 1 mol × S°(CaCO₃(s))
ΔS° = 1 mol × (39.75 J/K.mol) + 1 mol × (213.74 J/K.mol) - 1 mol × (92.9 J/K.mol)
ΔS° = 160.6 J/K. = 0.1606 kJ/K.
We can calculate the standard Gibbs free energy of the reaction (ΔG°) using the following expression.
ΔG° = ΔH° - T.ΔS°
where,
T: absolute temperature
<h3>A. 285 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 285K × 0.1606 kJ/K = 132.5 kJ
<h3>B. 1025 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1025K × 0.1606 kJ/K = 13.69 kJ
<h3>C. 1475 K</h3>
ΔG° = ΔH° - T.ΔS°
ΔG° = 178.3 kJ - 1475K × 0.1606 kJ/K = -58.59 kJ
