Density is defined as Mass per unit volume of an object ~
And hence, the correct choice is ~ 1
1 it a 2 it b 3 it c 4 it be 7 it a
Answer:
The standard change in free energy for the reaction = - 437.5 kj/mole
Explanation:
The standard change in free energy for the reaction:
4 KClO₃ (s) → 3 KClO₄(s) + KCl(s)
Given that ΔGf(KClO3(s)) = -290.9 kJ/mol;
ΔGf(KClO4(s)) = -300.4 kJ/mol;
ΔGf(KCl(s)) = -409 kJ/mol
According to Hess's law
ΔGr (Free energy change of reaction)= ∑(Product free energy - reactant free energy)
⇒ ΔGr⁰ = {3 x (-300.4) + (-409)} - {3 x (- 290.9)}
= - 901.2 - 409 + 872.7
= - 437.5 kj/mole
Answer:
6.4 L
Explanation:
When all other variables are held constant, you can use Boyle's Law to find the missing volume:
P₁V₁ = P₂V₂
In this equation, "P₁" and "V₁" represent the initial pressure and volume. "P₂" and "V₂" represent the final pressure and volume. You can find the theoretical volume by plugging the given values into the equation and simplifying.
P₁ = 3.2 atm P₂ = 1.0 atm
V₁ = 2.0 L V₂ = ? L
P₁V₁ = P₂V₂ <----- Boyle's Law
(3.2 atm)(2.0 L) = (1.0 atm)V₂ <----- Insert values
6.4 = (1.0 atm)V₂ <----- Simplify left side
6.4 = V₂ <----- Divide both sides by 1.0
The combustion reaction of propane would be expressed as:
C3H8 + 5O2 = 3CO2 + 4H2O
To determine the mass of water that is produced from the given amount of propane, we use the mass of propane and the relation of the substances from the balanced reaction. We do as follows:
moles propane = 22 g C3H8 ( 1 mol / 44.1 g ) = 0.50 mol C3H8
moles H2O = 0.50 mol C3H8 ( 4 mol H2O / 1 mol C3H8) = 2 mol H2O
mass H2O = 2 mol H2O ( 18.02 g / 1 mol ) = 36.04 g H2O
Therefore, the mass of water that is produced from 22 grams of propane would be 36.04 g.