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.
The answer to this would be 22 mL
NaCl would form because it’s a single replacement reaction
Answer:
pH of HNO₃ having an hydrogen ion concentration of 0.71M is 0.149
Explanation:
HNO₃ (aqueous) ⇄ H⁺ + NO3⁻
The pH is defined as the negative log of the hydrogen ion concentration
pH = - log [H⁺]
From the question, the hydrogen ion concentration is given as 0.71M, therefore
pH = -log [0.71]
= 0.149
Answer:
A. the jar covers
Explanation:
An independent variable is what you, as the scientist, changes.
