Answer is: 5.22·10²² atoms of Iodine.
m(CaI₂) = 12.75 g; mass of calcium iodide.
M(CaI₂) = 293.9 g/mol; molar mass of calcium iodide.
n(CaI₂) = m(CaI₂) ÷ M(CaI₂).
n(CaI₂) = 12.75 g ÷ 293.9 g/mol.
n(CaI₂) = 0.043 mol; amount of calcium iodide.
In one molecule of calcium iodide, there are two iodine atoms
n(I) = 2 · n(CaI₂).
n(I) = 0.086 mol; amount of iodine atoms.
Na = 6.022·10²³ 1/mol; Avogadro number.
N(I) = n(I) · Na.
N(I) = 0.086 mol · 6.022·10²³ 1/mol.
N(I) = 5.22·10²²; number of iodine atoms.
Answer:
A.
Explanation:
Water was added to the reaction after the completion of the reaction so as to lower the solubility if the product in the solution therefore, the product can be precipitated out. On adding water the reaction moves in forward direction and more product is formed. (By Le Chatelier's principle). Thus, the precipitation occurs. Hence, option A is correct.
Answer:
1) increase concentration
2) decrease the amount
3) decrease the concentration
4) it would increase
Explanation: edge 2021
Answer:
a)The mass percentages of nitrogen and phosphorus in the compound is 24.34% and 26.95% respectively.
b) 14.78 grams ammonia is incorporated into 100. g of compound.
Explanation:
a) Ammonium dihydrogen phosphate that is 
.
Molecular mass of ammonium dihydrogen phosphate = M
M = 115 g/mol
Percentage of an element in a compound:
Percentage of nitrogen:
Percentage of phosphorus:
b) Percentage of ammonia in 1 molecule of ammonium dihydrogen phosphate.
Molar mass of ammonia = 17 g/mol
Amount of ammonia in 100 grams of compound:
14.78% of 100 g of ammonium dihydrogen phosphate:
14.78 grams ammonia is incorporated into 100. g of compound.
Answer: A
0 degrees and 101 kPa are the conditions that describe the standard temperature and pressure. When expressed in K, the standard temperature 0 degrees equals 273.5 K. Also the standard pressure 101 kPa equals 760 mmHg or 1 Atm.
