Answer:
The answer to your question is 33.4 ml
Explanation:
Data
volume 1 = V1 = 42 ml
temperature 1 = T1 = 20°C
temperature 2 = T2 = -60°C
Volume 2 = V2 = x
Process
1.- Convert celsius to kelvin
T1 = 20 + 273 = 293°K
T2 = -60 + 273 = 233°K
2.- Use the Charles' law to solve this problem
Solve for V2
V2 = 
3.- Substitution
V2 = 
4.- Simplification
V2 = 
5.- Result
V2 = 33.4ml
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:
3 moles
Explanation:
To solve this problem we will use the Avogadro numbers.
The number 6.022×10²³ is called Avogadro number and it is the number of atoms, ions or molecules in one mole of substance. According to this,
1.008 g of hydrogen = 1 mole = 6.022×10²³ atoms.
18 g water = 1 mole = 6.022×10²³ molecules
we are given 36 g of C-12. So,
12 g of C-12 = 1 mole
24 g of C-12 = 2 mole
36 g of C-12 = 3 mole
So 3 moles of C-12 equals to the number of particles in 36 g of C-12.
