Answer: 40.1%
Explanation: The mass of calcium in this compound is equal to 40.1 grams because there's one atom of calcium present and calcium has an atomic mass of 40.1 . The molar mass of the compound is 100.1 grams. Using the handy equation above, we get: Mass percent = 40.1 g Ca⁄100.1 g CaCO3 × 100% = 40.1% Ca.
K, P, K, K, P, K, K, P, K, P. If it is moving, it is kinetic, if it isn't, it's potential. the sugar one is a little tricky using that method though, because we generally consider this in terms of spacial movement, but sugar holds energy which is later released by your body to allow you to move.the chemical bonds have potential energy because they release energy when broken.
Answer:
To prepare a 1 M solution, slowly add 1 formula weight of compound to a clean 1-L volumetric flask half filled with distilled or deionized water. Allow the compound to dissolve completely, swirling the flask gently if necessary.
Explanation:
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Answer:
23.2 g of Al will be left over when the reaction is complete
Explanation:
2Al + 3S → Al₂S₃
1 mol of Al = 26.98 g
1 mol of S = 32.06 g
Mole = Mass / Molar mass
63.8 g/ 26.98 g/m = 2.36 mole of Al
72.3 g / 32.06 g/m = 2.25 mole of S
2 mole of Aluminun react with 3 mole of sulfur
2.36 mole of Al react with (2.36 .3)/2 = 3.54 m of S
As I have 2.25 mole of S, and I need 3.54 S, is my limiting reagent so the limiting in excess is the Al.
3 mole of S react with 2 mole of Al
2.25 mole of S react with (2.25 m . 2)/3 = 1.50 mole
I need 1.50 mole of Al and I have 2.36, that's why the Al is in excess.
2.36 mole of Al - 1.50 mole of Al = 0.86 mole
This is the quantity of Al without reaction.
Molar mass . mole = Mass → 26.98 g/m . 0.86 m = 23.2 g
Answer:
Since you are producing 3.6 mol CO2, you can calculate the starting moles of CH4 with the simple mole-to-mole ratio: 1 mol CH4 / 1 mol CO2 as a conversion factor. Taking 3.6 mol CO2 x 1 mol CH4 / 1 mol CO2 = 3.6 mol CH4 (after canceling out the moles of CO2 on the top and bottom of the calculation)
Explanation:
