Let's divide the three experiments: The experiment with 10.00 mL of water is A), the experiment with 15.00 mL is B), and the experiment with 25.00 mL is C).
- (1) Now let's calculate the experimental density of each experiment. Density (ρ) is equal to the mass divided by the volume, thus:
- (2)To calculate the average density, we add each density and divide the result by the number of experiments (in this case 3):
- (3) The percent error is calculated by dividing the absolute value of the substraction of the theorethical and experimental values, by the theoretical value, times 100:
%error=
%error=
%error=2.44 %
Answer
0.9516 grams / mL (50.00 has 4 sig digs.)
Remark
You have a couple of extraneous numbers there. You don't care about anything except the mass of the flask + water/alcohol mixture (88.219 grams). and the mass of the flask (40.638 grams)
Formulas
- mass water/alcohol mixture = mass of the flask with fluid - mass flask
- density = mass / volume
Solution
mass water/alcohol mixture = 88.219 - 40.638 = 47.581
- Volume = 50 mL
- Density = mass / Volume
- Density = 47.581/50
- Density = 0.95162 There are 4 sig digs so the answer should be
- 0.9516
Answer: IT'S MUTATION I TOOK THE TEST ITS RIGHT HOPE THIS HELPED :)
Answer:
P1V1= P2V2
Explanation:
Inverse relationship
V2 = V1 X P1/P2
V2= 14.3 L x 45.0 mm Hg/63.0 mmHg= 8.99
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!
