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
Explanation:
Mass of the organic compound = 200g
Mass of carbon = 83.884g
Mass of hydrogen = 10.486g
Mass of oxygen = 18.640g
The mass of nitrogen = mass of organic compound - (mass of carbon + mass of hydrogen + mass of oxygen)
Mass of nitrogen = 200 - (83.884 + 10.486 + 18.64) = 200 - 113.01
Mass of nitrogen = 86.99g
The empirical formula of a compound is its simplest formula.
It is derived as shown below;
C H O N
Mass 83.884 10.486 18.64 86.99
molar
mass 12 1 16 14
Moles 83.884/12 10.486/1 18.64/16 86.99/14
6.99 10.49 1.17 6.21
Divide
by
lowest 6.99/1.17 10.49/1.17 1.17/1.17 6.21/1.17
6 9 1 5
Empirical formula C₆H₉ON₅
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Answer:
-372000 J or -372 KJ
Explanation:
We have the electrochemical reaction as;
Mg(s) + Fe^2+(aq)→ Mg^2+(aq) + Fe(s)
We must first calculate the E∘cell from;
E∘cathode - E∘anode
E∘cathode = -0.44 V
E∘anode = -2.37 V
Hence;
E∘cell = -0.44 V -(-2.37 V)
E∘cell = 1.93 V
n= 2 since two electrons were transferred
F=96,500C/(mol e−)
ΔG∘=−nFE∘
ΔG∘= -( 2 * 96,500 * 1.93)
ΔG∘= -372000 J or -372 KJ
Aluminum has three oxidation states. The most common one is +3. The other two are +1 and +2. One +3 oxidation state for Aluminum can be found in the compound aluminum oxide, Al2O3.
The blank is atoms. Structural formulas show the atomic arrangement in a molecule.
