The given thermochemical reaction is between hydrogen gas and chlorine gas to form hydrogen chloride.
This can be represented as:
Δ
=-184.6 kJ/mol
So when two moles of HCl is formed, 184.6 kJ of energy is released.
Calculating the heat released when 3.18 mol HCl (g) is formed in the reaction:
Therefore, 293.5 kJ of heat is released when 3.18 mol HCl is formed in the reaction between hydrogen and chlorine.
Answer:
see explanation below
Explanation:
To do this exercise, we need to use the following expression:
P = nRT/V
This is the equation for an ideal gas. so, we have the temperature of 22 °C, R is the gas constant which is 0.082 L atm / mol K, V is the volume in this case, 5 L, and n is the moles, which we do not have, but we can calculate it.
For the case of the oxygen (AW = 16 g/mol):
n = 30.6 / 32 = 0.956 moles
For the case of helium (AW = 4 g/mol)_
n = 15.2 / 4 = 3.8 moles
Now that we have the moles, let's calculate the pressures:
P1 = 0.956 * 0.082 * 295 / 5
P1 = 4.63 atm
P2 = 3.8 * 0.082 * 295 / 5
P2 = 18.38 atm
Finally the total pressure:
Pt = 4.63 + 18.38
Pt = 23.01 atm
Answer:
1.6 L
Explanation:
Using Charle's law
Given ,
V₁ = 1.5 L
V₂ = ?
T₁ = 12 °C
T₂ = 32 °C
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
So,
T₁ = (12 + 273.15) K = 285.15 K
T₂ = (32 + 273.15) K = 305.15 K
Using above equation as:
New volume = 1.6 L
The reaction is given as:
Here, two moles of copper nitrate reacts with four moles of potassium iodide to give two moles of copper iodide, one mole of iodine and four moles of potassium nitrate.
First, calculate the number of moles of copper nitrate.
Number of moles is equal to the product of molarity and volume of solution in litre.
Number of moles = 
(1 L =1000 mL)
= 
Copper nitrate requires = 
mole of potassium iodide
= 
of potassium iodide
Volume of solution in litre = 
Thus, volume of potassium iodide is =
= 
1 L =1000 mL
Volume of potassium iodide in mL =
Hence, 0.2089 M potassium iodide consist of sufficient potassium iodide to react with copper nitrate in 3.88 mL of a 0.3842 M solution of copper nitrate .
