gThe atomic radii of a divalent cation and a monovalent anion are 0.085 nm and 0.125 nm, respectively. (a) Calculate the force o
1 answer:
Answer:
1.05 × 10⁻⁸ N
Explanation:
Using Coulomb's law
F force of attraction between them = k = k
K coulomb constant = 8.99 × 10⁹ N m²C⁻²
Z₁ = + 2 ( valency electron)
Z₂ = -1 ( valency electron)
r is the distance between then since they just touch one another = 0.085 nm + 0.125 nm = 0.21 nm = 0.21 × 10⁻⁹ m
q charge = 1.602 × 10⁻¹⁹ C of an electron
F = ( 2 × 1 × 8.99 × 10⁹ N m²C⁻² × (1.602 × 10⁻¹⁹ C)² / ( 0.21 × 10⁻⁹ m )² = 1046.34 × 10 ⁻¹¹ = 1.05 × 10⁻⁸ N
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Answer:
- 13.56 g of sodium chloride are theoretically yielded.
- Limiting reactant is copper (II) chloride and excess reactant is sodium nitrate.
- 0.50 g of sodium nitrate remain when the reaction stops.
- 92.9 % is the percent yield.
Explanation:
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In this case, according to the question, it is possible to set up the following chemical reaction:
Thus, we can first identify the limiting reactant by computing the yielded mass of sodium chloride, NaCl, by each reactant via stoichiometry:
Thus, we infer that copper (II) chloride is the limiting reactant as it yields the fewest grams of sodium chloride product. Moreover the formed grams of this product are 13.56 g. Then, we take 13.56 g of sodium chloride to compute the consumed mass sodium nitrate as it is in excess:
Therefore, the leftover of sodium nitrate is:
Finally, the percent yield is computed via:
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