Answer:
It is easy
Explanation:
give me BRainlist Answer and I will answer ( a piece of cake)
Answer: The answer is 167
Explanation: This is because that was right on edg. so yea heart this tho plsss
Answer:
1750L
Explanation:
Given
Initial Temperature = 25°C
Initial Pressure = 175 atm
Initial Volume = 10.0L
Final Temperature = 25°C
Final Pressure = 1 atm
Final Volume = ?
This question is an illustration of ideal gas law.
From the given parameters, the initial temperature and final temperature are the same; this implies that the system has a constant temperature.
As such, we'll make use of Boyle's Law to solve this;
Boyle's Law States that:
P₁V₁ = P₂V₂
Where P₁ and P₂ represent Initial and Final Pressure, respectively
While V₁ and V₂ represent Initial and final volume
The equation becomes
175 atm * 10L = 1 atm * V₂
1750 atm L = 1 atm * V₂
1750 L = V₂
Hence, the final volume that can be stored is 1750L
Answer:
D.
Explanation:
Water is polar, for one thing. Polar mixes with polar, nonpolar mixes wih nonpolar. This leaves D.
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!
