Answer:
P(total) = 1110 mmHg
Explanation:
According to the Dalton law of partial pressure,
The pressure exerted by mixture of gases are equal to the sum of partial pressure of individual gases.
P(total) = P1 + P2 + P3+ .....+ Pn
Given data:
Sample A = 740 mmHg
Sample B = 740 mmHg
Sample C = 740 mmHg
Total pressure = ?
Solution:
<em>Sample A:</em>
P₁V₁ = P₂V₂
P₂ = P₁V₁ / V₂
P₂ = 740 mmHg × 2L/4L
P₂ = 370 mmHg
<em>Sample B:</em>
P₁V₁ = P₂V₂
P₂ = P₁V₁ / V₂
P₂ = 740 mmHg × 2L/4L
P₂ = 370 mmHg
<em>Sample C:</em>
P₁V₁ = P₂V₂
P₂ = P₁V₁ / V₂
P₂ = 740 mmHg × 2L/4L
P₂ = 370 mmHg
Total pressure:
P(total) = P1 + P2 + P3
P(total) = 370 mmHg + 370 mmHg+ 370 mmHg
P(total) = 1110 mmHg
Dilution<span> is when you decrease the concentration of a </span>solution<span> by adding a solvent. As a result, if you want to </span>dilute<span> salt water, just add water. ... Add more solute until it quits dissolving. That point at which a solute quits dissolving is the point at which it's </span>saturated<span>.</span>
Explanation :
In the given case different law related to gas is given. The attached figure shows the required solution.
Boyle's law states that the pressure is inversely proportional to the volume of the gas i.e.


k is a constant.
Charle's law states that the volume of directly proportional to the temperature of the gas.


Combined gas law is the combination of the pressure, volume and the temperature of the gas i.e.

Hence, this is the required solution.
The balanced chemical reaction is written as:
4Al + 3O2 = 2Al2O3
To determine the mass of oxygen gas that would react with the given amount of aluminum metal, we use the initial amount and relate this amount to the ratio of the substances from the chemical reaction. We do as follows:
moles Al = 16.4 g ( 1 mol / 26.98 g ) = 0.61 mol Al
moles O2 = 0.61 mol Al ( 3 mol O2 / 4 mol Al ) = 0.46 mol O2
mass O2 = 0.46 mol O2 ( 32.0 g / mol ) = 14.59 g O2
Therefore, to completely react 16.4 grams of aluminum metal we need a minimum of 14.59 grams of oxygen gas.
Answer:
[C] carbon solid
Explanation:
Pure solids and liquids are never included in the equilibrium constant expression because they do not affect the reactant amount at equilibrium in the reaction, thus since your equation has [C] as solid it will not be part of the equlibrium equation.