Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
M(Ca(NO3)2)= M(Ca) + M(N) + 6M(O)= 40.0 +14.0 +6*16.0 = 150 g/mol
15.0 g Ca(NO3)2 * 1mol/150 g = 0. 100 mol Ca(NO3)2
Ca(NO3)2 -------> Ca²⁺ +2NO3⁻
1 mol 2 mol
0.100 mol 0.200 mol
We have 0.2 mol NO3⁻ in 300. mL=0.300 L of solution,
so
0.200 mol NO3⁻ / 0.300 L solution ≈ 0.667 mol NO3⁻ /L solution = 0.667 M
Concentration of NO3⁻ is 0.667 M.
Review and Study Material Before Going to
Class.
Seek Understanding.
Take Good Notes.
Practice Daily.
Take Advantage of Lab Time.
Use Flashcards.
Use Study Groups.
Break Large Tasks Into Smaller Ones.
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Answer;
-The charges of the positive and negative copper ions cancel each other out.
Explanation;
-A normal atom has a neutral charge with equal numbers of positive and negative particles.That means an atom with a neutral charge is one where the number of electrons is equal to the atomic number.
-Atoms are electrically neutral because they have equal numbers of protons (positively charged) and electrons (negatively charged). If an atom gains or loses one or more electrons, it becomes an ion. If it gains one or more electrons, it now carries a net negative charge, and is thus anionic.
Take a hypothetical sample of exactly 100 grams of the solution.
(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea
((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O
(0.2664 mol) /0.0840 (kg) = 3.17143mol/kg = 3.18m urea
