Answer:
The mass percent of aluminum sulfate in the sample is 16.18%.
Explanation:
Mass of the sample = 1.45 g
Mass of the precipitate = 0.107 g
Moles of aluminum hydroxide = 
According to reaction, 2 moles of aluminum hydroxide is obtained from 1 mole of aluminum sulfate .
Then 0.001372 moles of aluminum hydroxide will be obtained from:
Mass of 0.000686 moles of aluminum sulfate :
= 0.000686 mol × 342 g/mol = 0.2346 g
The mass percent of aluminum sulfate in the sample:
The HONC 1234 rule is a way to remember the bonding tendencies of hydrogen, oxygen, nitrogen, and carbon atoms in molecules. Hydrogen tends to form one bond, oxygen two, nitrogen three and carbon four.
<span>17.5 g
35 ppt stands for 35 parts per thousand. So let's convert that to a decimal number by taking 35 and dividing by 1000.
35/1000 = 0.035
Now multiply that number by the number of grams of seawater you have. So
0.035 * 500 g = 17.5 g
So you have 17.5 grams of salt when you have 500 grams of seawater.</span>
Answer:
145.8g
Explanation:
Given parameters:
Number of moles of magnesium hydroxide = 2.5mol
Unknown:
Mass of Mg(OH)₂ = ?
Solution:
To solve this problem we use the expression below;
Mass of Mg(OH)₂ = number of moles x molar mass
Molar mass of Mg(OH)₂ = 24.3 + 2(16 + 1) = 58.3g/mol
Mass of Mg(OH)₂ = 2.5 x 58.3 = 145.8g
I used an online calculator and got 54.22 m/s. I hope that helps
