Answer:
219.78 mL of the stock solution are needed
Explanation:
First, we take a look at the desired Al2(SO4)3 working solution. We are told that the we need 400 mL of an aqueous aluminum sulfate solution 1.0 M. Let's see how many moles of the compound we have in the desired volume:
1000 mL Al2(SO4)3 solution ----- 1 mole of Al2(SO4)3
400 mL Al2(SO4)3 solution ----- x = 0.4 moles of Al2(SO4)3
To reach the desired concentration in the working solution we need 0.4 moles of Al2(SO4)3 in 400 mL, so we calculate the volume of the stock solution needed to prepare the working solution:
1.82 moles Al2(SO4)3/L = 1.82 M → This is the molar concentration of the stock solution.
1.82 moles of Al2(SO4)3 ----- 1000 mL
0.4 moles of Al2(SO4)3 ----- x = 219.78 mL
So, if we take 219.78 mL of the 1.82 M stock solution, we put it in a graduated cylinder and we dilute it to 400 mL, we would obtain a 1.0 M Al2(SO4)3 solution.
Answer:
You kinda left out the atom being discussed.
But My guts tell me its Berrylium
Answer: "Beryllium – 10".
As per as the Multiplication rules of the significant figures, whenever any numbers in the decimals forms are multiplied or divided then result in mentioned in such a way so that the significant figures after the decimal will be same as that in the given least condition.
_______________________________
102900/12 = 8575
170 × 1.27 = 215.9
∴ (102,900 ÷ 12) + (170 × 1.27) = 8575 + 215.9
= 8790.9
Now, As per as Above rules, answer in correct significant figures will be = 8791.
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!
