First, we need to assume acetylene to adapt an ideal gas behavior. So, we can use the Ideal Gas Law:
PV = nRT
Given:
V = 42 L
T = 305 K
P = 780 torr = 1.026 atm
R = gas constant, 0.0821 L atm/ mol K
n = amount of gas in the canister
1.026 atm * 42 L = n * 0.0821 L atm/mol K * 305 K
n = 1.72 moles of gas
Answer:
7.5 mol
Step-by-step explanation:
2Al + 3Br₂ ⟶ 2AlBr₃
You want to convert moles of AlBr₃ to moles of Br₂.
The molar ratio is 3 mol Br₂:2 mol AlBr₃.
Moles of Br₂ = 5 × 3/2
Moles of Br₂ = 7.5 mol Br₂
You need 7.5 mol of Br₂ to produce 5 mol of AlBr₃.
Answer:
The answer is B on edge
Explanation:
Here are my notes on this section for anyone that needs them
Enthalpy and State Function
Bonds contain potential energy. Breaking and forming bonds involves energy. Reactants and products contain energy. Enthalpy (H) is a measure of heat and internal energy in a system.
A state function is a quantity whose change in magnitude during a process depends only on the beginning and end points the process, not the path taken between them. Enthalpy change during reaction depends only on the identity of reactants and products and their initial and finial condition
Enthalpy of Formation
enthalpy of formation (Hf) is the energy absorbed or released when a pure substance forms from elements in their standard states
Units: kJ/mol, kcal/mol
Standard state is the natural state of an element at 1 atm (atmosphere of pressure) and 25 degrees celsius. Hf for a pure element in its standard state is 0 kJ/mol.
H (hydrogen): H2(g)
N (nitrogen): N2(g)
O (oxygen): O2(g)
F (fluorine): F2(g)
Cl (chlorine): Cl2(g)
Br (bromine): Br2(l)
Hg (mercury): Hg(l)
Enthalpy of Reaction
Enthalpy of reaction (Hrxn) is energy absorbed or released during a chemical reaction
Hrxn negative: exothermic reaction
Hrxn positive: endothermic reaction
Hess's Law: Hrxn = Σ(ΔHƒ, products) − Σ(ΔHƒ, reactants)
thermochemical equation: the chemical equation that shows the state of each substance involved and the energy change involved in a reaction
Find the kJ/mol of the product and then subtract the kJ/mol of the reactants.
<u>Answer:</u> The experimental van't Hoff factor is 1.21
<u>Explanation:</u>
The expression for the depression in freezing point is given as:
where,
i = van't Hoff factor = ?
= depression in freezing point = 0.225°C
= Cryoscopic constant = 1.86°C/m
m = molality of the solution = 0.100 m
Putting values in above equation, we get:
Hence, the experimental van't Hoff factor is 1.21
