2
The answer is 1 527 mg aluminium.
Amount of sodium sulfate required = 0.683 x 350/100 = 0.239The formula mass of sodium sulfate, Na2SO4 is 2(23.0) + 32.1 + 4(16.0) = 142.1 Therefore mass of sodium sulfate required = 0.239 x 142.1 = 34.0g Approximately 34 grams of sodium sulfate would be needed.
63.3 g
44.1-44.5 -Apex
3 grams
4,12 grams aluminum sulfate is equivalent to 0,012 moles (for the anhydrous salt).
The answer is 1 527 mg aluminium.
Aluminium Sulphate is Al2(SO4)3 and its mass is27x2 + 96x3 = 342g/mole
Amount of sodium sulfate required = 0.683 x 350/100 = 0.239The formula mass of sodium sulfate, Na2SO4 is 2(23.0) + 32.1 + 4(16.0) = 142.1 Therefore mass of sodium sulfate required = 0.239 x 142.1 = 34.0g Approximately 34 grams of sodium sulfate would be needed.
g
To find this answer it is necessary to first find the chemical formula of aluminum sulfate utilizing the valence charges from the periodic table. Aluminum Sulfate is a molecular formed by the ionic bonding of aluminum to to the polyatomic ion of sulfate Al = +3 SO4 is a polyatomic ion with a charge of -2 Therefore after balancing the equation, you get the chemical formula Al2(SO4)3 Next, in order to find osmolarity, you must first find molarity. This is done by converting grams to moles. Moles are calculated in grams/Liter and is equal to the molecular weight. What is the molecular weight of aluminum sulfate. Looking at the periodic table, you sum the atomic weights of each element, shown in AMUs. Al = 27 (x2) = 54 S = 32 (x3) =96 O = 16 (x12) = 192 We add these up to determine the molecular weight of the substance, which can be used to determine the molarity of this solution. MW = 342g, which means that one mole of aluminum sulfate is 342 grams per liter of water. We can now use the information presented in the above problem to determine molarity 10 grams dissolved in 200 ml of water. Since measurements are in grams per liter, we can assume that if 10 grams were dissolved in 200 ml of water, then 50 grams will dissolve in 1000 ml, or 50 grams per liter (multiply both factors by five). We must use our grams of solute to convert grams into our molarity 50 g x 1 mole/342 grams of aluminum sulfate to determine the molarity. = .15 M At this point, to determine osmolarity, you must be aware that osmolarity is the measure of PARTICLES within a solution per liter. Realizing that aluminum sulfate is held together by an ionic bond, it will freely dissolve in water and produce Al ions and SO4 ions. So we sum the particles of each: 2 Al ions per molecule 3 SO4 ions per molecule Therefore, multiplying the molarity of the aluminum sulfate solution times 5 (total 5 particles from each molecule) will give you the osmolarity. = 0.73 Osm per liter of aqueous solution.
63.3 g
800 g oxygen are needed.
44.1-44.5 -Apex
3 grams
211 is the mass in grams of 1.15 moles of strontium sulfate.
33 grams aluminum (1 mole Al/26.98 grams) = 1.2 moles of aluminum ================