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
Hey there!:
Molar mass of Mg(OH)2 = 58.33 g/mol
number of moles Mg(OH)2 :
moles of Mg(OH)2 = 30.6 / 58.33 => 0.5246 moles
Molar mass of H3PO4 = 97.99 g/mol
number of moles H3PO4:
moles of Mg(OH)2 = 63.6 / 97.99 => 0.649 moles
Balanced chemical equation is:
3 Mg(OH)2 + 2 H3PO4 ---> Mg3(PO4)2 + 6 H2O
3 mol of Mg(OH)2 reacts with 2 mol of H3PO4 ,for 0.5246 moles of Mg(OH)2, 0.3498 moles of H3PO4 is required , but we have 0.649 moles of H3PO4, so, Mg(OH)2 is limiting reagent !
Now , we will use Mg(OH)2 in further calculation .
Molar mass of Mg3(PO4)2 = 262.87 g/mol
According to balanced equation :
mol of Mg3(PO4)2 formed = (1/3)* moles of Mg(OH)2
= (1/3)*0.5246
= 0.1749 moles of Mg3(PO4)2
use :
mass of Mg3(PO4)2 = number of mol * molar mass
= 0.1749 * 262.87
= 46 g of Mg3(PO4)2
Therefore:
% yield = actual mass * 100 / theoretical mass
% = 34.7 * 100 / 46
% = 3470 / 46
= 75.5%
Hope that helps!
Answer:
Changes in Properties Changes in properties result when new substances form. For instance, gas production, formation of a precipitate, and a color change are all possible evidence that a chemical reaction has taken place. ... Change in Color A color change can signal that a new substance has formed.
Explanation:
Answer : The new volume of the air is, 6.83 L
Explanation :
Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
where,
are the initial volume and temperature of the gas.
are the final volume and temperature of the gas.
We are given:
Putting values in above equation, we get:
Therefore, the new volume of the air is, 6.83 L
