Answer:
The enthalpy is increased by the increased heat of the reaction.
Explanation:
In this reaction, as the transition from solid to liquid state, enthalpy increases, that is, the heat applied to change its state is greater and this increases, reaching a mayor disorder.
If the reaction increases its heat, and a liquid state passes, the enthalpy increases, therefore the disorder also and the entropy will also be increased.
Answer:
3.5 atm
Explanation:
As stated in the question pressure is required to counteract the natural tendency for water to dilute the more concentrated solution. The difference in concentrations will give us the answer using the osmotic pressure equation.
π = ( n/v) RT where n/v is the molarity (mol/L), R is the gas constant and T is the temperature.
The difference in osmotic pressure of the solutions is:
Δπ = Δ c RT where c is the difference in molar concentrations.
pressure required = Δπ = (0.190 - 0.048) M x 0.821 Latm/Kmol x 298 K
= 3.47 atm
The balanced equation for the above reaction is as follows;
2S + 3O₂ --> 2SO₃
Stoichiometry of O₂ to SO₃ is 3:2
O₂ is the limiting reactant and S is provided in excess. since O₂ is the limiting reactant, the whole amount is consumed in the reaction and amount of product formed depends on amount of limiting reactant present.
Number of O₂ moles reacted- 4 g / 32 g/mol = 0.125 mol
3 mol of O₂ forms 2 mol of SO₃
therefore when 0.125 mol of O₂ reacts number of SO₃ moles - 2/3 x 0.125 mol
Number of SO₃ moles formed - 0.0833 mol
Answer is 4) 0.08 mol
Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is
To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
