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Tomtit [17]
3 years ago
15

Mannoheptulose is a sugar found in avocados. If each C-C bond contains 76 kcal of energy and each C-H bond contains 91 kcal, how

many kcal of energy are available through the complete catabolism of this sugar

Chemistry
1 answer:
nadya68 [22]3 years ago
6 0

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

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An electrochemical cell is constructed using two half-cells: Al(s) in Al(NO2)3(aq) and Cu(s) in Cu(NO3)2(aq). The two half cells
wlad13 [49]

Answer:

Cu(s) in Cu(NO₃)₂(aq)

Explanation:

The standard reduction potential (E°) is the energy necessary to reduce the atom in a redox reaction. When an atom reduces it gains electrons from other than oxides. As higher is E°, easily it will reduce. The substance that reduces is at the cathode of a cell, where the electrons go to, and the other that oxides are at the anode of the cell.

The standard reduction potentials from Al(s) and Cu(s) are, respectively, -1.66V and +0.15V, so the half-cell of Cu(s) in Cu(NO₃)₂(aq) is the cathode.

4 0
3 years ago
C. Use Hess's law and the following equations to calculate ΔH for the reaction 4NH3 (g) + 5O2 (g) 4NO(g) + 6H2 O(g). Show your w
Monica [59]

Considering the Hess's Law, the enthalpy change for the reaction is -906.4 kJ/mol.

<h3>Hess's Law</h3>

Hess's Law indicates that the enthalpy change in a chemical reaction will be the same whether it occurs in a single stage or in several stages. That is, the sum of the ∆H of each stage of the reaction will give us a value equal to the ∆H of the reaction when it occurs in a single stage.

<h3>ΔH in this case</h3>

In this case you want to calculate the enthalpy change of:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O

which occurs in three stages.

You know the following reactions, with their corresponding enthalpies:

Equation 1: 2 N₂ + 6 H₂ → 4 NH₃   ΔH = –183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 2 H₂ + O₂→ 2 H₂O     ΔH = -483.7 kJ/mol

Because of the way formation reactions are defined, any chemical reaction can be written as a combination of formation reactions, some going forward and some going back.

In this case, first, to obtain the enthalpy of the desired chemical reaction you need 4 moles of NH₃ on reactant side and it is present in first equation on product side. So you need to invert the reaction, and when an equation is inverted, the sign of delta H also changes.

Now, 4 moles of NO must be a product and is present in the second equation, so let's write this as such.

Finally, you need 6 moles of H₂O on the product side, so you need to multiply by 3 the third equation to obtain the amount of water that you need. Since enthalpy is an extensive property, that is, it depends on the amount of matter present, since the equation is multiply by 3, the variation of enthalpy also.

In summary, you know that three equations with their corresponding enthalpies are:

Equation 1: 2 4 NH₃ → N₂ + 6 H₂  ΔH = 183.6 kJ/mol

Equation 2:  2 N₂ + 2 O₂ → 4 NO     ΔH = 361.1 kJ/mol

Equation 3: 6 H₂ + 3 O₂→ 6 H₂O     ΔH = -1,451.1 kJ/mol

Adding or canceling the reactants and products as appropriate, and adding the enthalpies algebraically, you obtain:

4 NH₃ + 5 O₂ → 4 NO + 6 H₂O ΔH= -906.4 kJ/mol

Finally, the enthalpy change for the reaction is -906.4 kJ/mol.

Learn more about Hess's law:

brainly.com/question/5976752

brainly.com/question/13707449

brainly.com/question/13707449

brainly.com/question/6263007

brainly.com/question/14641878

brainly.com/question/2912965

#SPJ1

7 0
2 years ago
For the reaction at 298 K, 2NO2(g) N2O4(g) the values of ΔH° and ΔS° are -58.03 kJ and -176.6 J/K, respectively. Calculate the v
Ierofanga [76]

Answer:

\Delta G^o=-5.4032 kJ

The temperature for \Delta G^o=0[/tex is [tex]T=328.6 K

Explanation:

The three thermodinamic properties (enthalpy, entropy and Gibbs's energy) are linked in the following formula:

\Delta G^o=\Delta H^o + T*\Delta S^o

Where:

\Delta G^o is Gibbs's energy in kJ

\Delta H^o is the enthalpy in kJ

\Delta S^o is the entropy in kJ/K

T is the temperature in K

Solving:

\Delta G^o=-58.03 kJ - 298K*-0.1766 kJ/K

\Delta G^o=-5.4032 kJ

For \Delta G^o=0:

0=\Delta H^o - T*\Delta S^o

\Delta H^o= T*\Delta S^o

T=\frac{\Delta H^o}{\Delta S^o}

T=\frac{-58.03 kJ}{-0.1766 kJ/K}

T=328.6 K

3 0
3 years ago
Read 2 more answers
rutherford's experiment indicated that matter was not as uniform as it appears what part of his experimental results implied thi
MaRussiya [10]

Rutherford theorized that atoms have their charge concentrated in a very small nucleus.

This was famous Rutherford's Gold Foil Experiment: he bombarded thin foil of gold with positive alpha particles (helium atom particles, consist of two protons and two neutrons).  

Rutherford observed the deflection of alpha particles on the photographic film and notice that most of alpha particles passed straight through foil.

That is different from Plum Pudding model, because it shows that most of the atom is empty space.

According to Rutherford model of the atom:

1) Atoms have their charge concentrated in a very small nucleus.

2) Major space in an atom is empty.

3) Atoms nucleus is surrounded by negatively charged particles called electrons.  

4) An atom is electrically neutral.

6 0
3 years ago
In the chemical formula 3nh3, how many atoms of hydrogen (h) are there? 3 6 9 11
ANTONII [103]
Your answer is 3! Hope you get a good grade :)♥
7 0
3 years ago
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