Convert 55.0g Ca(OH)2 to moles.
The calculation shows that 55.0g of Ca(OH)2 corresponds to 0.742 moles.
To find the volume, divide 0.742 mol of Ca(OH)2 by its molarity of 0.680M, yielding approximately 1.09L of Ca(OH)2.
If you disregard the negligible volume of the Ca(OH)2 itself, the resulting total volume of a 0.680M solution created by dissolving 55.0g of Ca(OH)2 in an appropriate amount of water would be 1.09L.
Answer:
Explanation:
1) Alkali metals and halogens both need to achieve a stable outer electron shell, requiring alkali metals to lose one electron and halogens to gain one.
2) They share an identical count of outer shell electrons.
3) Typically, they have elevated melting points.
4) They exhibit low reactivity or none at all.
5) They belong to group 7.
It must be a physical change unless a chemical substance interacts with the metal surface, resulting in a chemical change.
The ore contains a 55.4% composition of calcium phosphate (related to apatite), leading to a calculation where Ca3(PO4)2 equals 55.4%x=1000g, resulting in x=1000/0.554, which equals 1.805kg. To find the percentage of phosphorus in this amount of calcium phosphate, calculate the total masses of the elements in Ca3PO4= Ca=40.078 x 3= 120.23 and (PO4)2= (30.974+64)2=189.95 (noting that oxygen contributes 16 mass x 4 =64), giving a cumulative mass of 310.2, while the phosphorus is 61.95 (Pmass x 2). Therefore, 61.95/310.2= 0.19 or 19% for phosphorus. Consequently, from 1.805 x 0.19, there will be 0.34kg of phosphorus.
