The chemical equation is said to be balanced if the number of atoms in the reactants and products is the same
<h3>Further explanation</h3>
Equation balanced ⇒ total number of atoms in reactants(on the left)= total number of atoms in products(on the right)
H₂+O₂---> H₂O
Reactants : H₂, O₂
Products : H₂O
not balanced
H₂O₂ ---> H₂O+O₂
Reactants : H₂O₂
Products : H₂O, O₂
not balanced
Na+O₂ ---> Na₂O
Reactants : Na, O₂
Products : Na₂O
not balanced
N₂+H₂ ---> NH₃
Reactants : N₂, H₂
Products : NH₃
not balanced
P₄+O₂---> P₄O₁₀
Reactants : P₄, O₂
Products : P₄O₁₀
not balanced
Fe+H₂O ----> Fe₃O₄ + H₂
Reactants : Fe, H₂O
Products : Fe₃O₄
not balanced
The term that refers to compounds that can form hydrates but do not contain water molecules is anhydrous.
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)
=> p1 = 2 p2
Which is easy to demonstrate using ideal gas equation:
p1 = nRT/V = 2.0 mol * RT / 1 liter
p2 = nRT/V = 1.0 mol * RT / 1 liter
=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2
2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.
So, the pressure in both chambers (which form one same vessel) is:
p = nRT/V = 3.0 mol * RT / 2liter
which compared to the initial pressure in chamber 1, p1, is:
p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1
So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.
You can also see how the pressure in chamber 2 changes:
p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
