Consider the following reaction: Solid zinc was added to 1.0 M HC1.
1 answer:
You might be interested in
Answer:
- The first picture attached is the diagram that accompanies the question.
- The<u> second picture attached</u> is the diagram with the answer.
Explanation:
In the box on the left there are 8 Cl⁻ ions and 8 Na⁺ ions.
The dissociaton equation for NaCl(aq) is:
- NaCl (aq) → Na⁺ (aq) + Cl⁻(aq)
The dissociation equation for CaCl₂ (aq) is:
- CaCl₂ (aq) → Ca²⁺ (aq) + 2Cl⁻(aq)
A 0.10MCaCl₂ (aq) solution will have half the number of CaCl₂ units as the number of NaCl units in a 0.20M NaCl (aq) solution.
Thus, while the 0.20M NaCl (aq) solution yields 8 ions of Na⁺ and 8 ions of Cl⁻, the 0.10MCaCl₂ (aq) solution will yield 4 ions of Ca²⁺ (half because the concentration if half) and 8 ions of Cl⁻ (first take half and then multiply by 2 because the dissociation reaction).
Thus, your drawing must show 4 dots representing Ca²⁺ ions and 8 dots representing Cl⁻ ions in the box on the right.
Answer:
Total Kcal energy produced in the catabolism of mannoheptulose = 1184 Kcal
Explanation:
The molecular formula of mannoheptulose is C₇H₁₄O₇.
The structure is as shown in the attachment below.
Number of C-C bonds present in mannoheptulose = 6
Number of C-H bonds present in mannoheptulose = 8
Since the each C-C bond contains 76 Kcal of energy,
Amount of energy present in six C-C bonds = 6 * 76 = 456 Kcal
Also, since each C-H bond contains 91 Kcal of energy;
amount of energy present in eight C-H bonds = 8 * 91 = 728 Kcal
Total Kcal energy produced in the catabolism of mannoheptulose = 456 + 728 = 1184 Kcal
