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Paraphin [41]
3 years ago
11

Carbohydrate metabolic enzymes bind D-glucose specifically. D-glucose has an estimated caloric value of 1 k.cal per 4 grams of c

arbohydrate.
Based on the methods used to convert the energy of D-glucose into a useable form, what would you estimate the caloric value of L-glucose to be using the same method?

A) 1 k.cal per 4 grams
B) 1 k.cal per 2 grams
C) 1 k.cal per 8 grams
D) 0 k.cal per 4 grams
E) 2 k.cal per 8 grams
Chemistry
1 answer:
Aleksandr [31]3 years ago
3 0

Answer:

D) 0 k.cal per 4 grams

Explanation:

Genarally, glucose can be classified into two enantiomers such as d-glucose and l-glucose. The d-glucose is the most common sugar that bodies of living organisms use as source of energy. However, l-glucose is an organic compound and it is one of the aldohexose monosaccharides. It is the l-isomer of glucose and commonly refer to as a low-calorie sweetener. The l-glucose is  relatively indistinguishable in taste from d-glucose but cannot be used as a source of energy. Therefore, in the given problem:

if d-glucose has an estimated caloric value of 1 k.cal per 4 grams of carbohydrate, then the caloric value of l-glucose will be 0 k.cal per 4 grams.

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Calculate the formal charges of the atoms in co co2 and co32-
Dimas [21]
Every element in the compound has its own formal charge. The formula is:
Formal charge = Valence electrons - Nonbonding electrons - Bonding electrons/2
Let the Lewis structure of the compound aid you.

1. For CO:
Formal charge of C = 4 - 2 - 6/2 = -1
Formal charge of O = 6 - 2 - 6/2 = +1

2. For CO₂:
Formal charge of C = 4 - 0 - 8/2 = 0
Formal charge of each O = 6 - 4 - 4/2 = 0

3. For CO₃²⁻:
Formal charge of C: 4 - 0 - 8/2 = 0
Formal charge of two O's single-bonded to C = 6 - 6 - 2/2 = -1
Formal charge of the O double-bonded to C = 6 - 4 - 4/2 = 0

6 0
4 years ago
The following information is given for ether, C2H5OC2H5, at 1atm: boiling point = 34.6 °C Hvap(34.6 °C) = 26.5 kJ/mol specific h
Likurg_2 [28]

Answer:

The heat is  H= -8.044KJ

Explanation:

From the question we are told that

   The pressure is  P = 1 \ atm

    The boiling point is  B_P  = 34^oC

     The heat of vaporization at 34°C is  = 26.5 kJ/mol

       The specific heat of the liquid is c_p = 2.32 J/g^oC

      The  mass is  m = 22.5g

  The no of moles of the sample of C_2 H_5OC_2H_5  is given as

               No \ mole (n)  = \frac{Mass \ of  \ sample }{Molar \ mass }

    The  Molar mass for C_2 H_5OC_2H_5  is a value = = 74.12 g/mol

Substituting the value into the above equation

          n  = \frac{22.5}{74.12}

                           = 0.30356 \ mol

      The heat H is mathematically as

                H  =- nH_{vap}

The negative sign show that the heat is for condensing

            H = 0.30356 * 26.5 *10^{3} J/mol

                H= -8.044KJ

                   

6 0
4 years ago
What are the possible values of n and ml for an electron in a 5d orbital?
liberstina [14]

Answer:

The answer to your question is n = 5, l = 2, m can be -2, -1, 0, 1 or 2

Explanation:

Data

orbital = 5d

values of n, l, m

Process

1.- Determine the value of n

n is the coefficient of the orbital, in this problem n = 5

2.- Determine the value of l

l takes values depending in the sublevel of energy,

if the sublevel is s  then l = 0

                           p          l = 1

                           d          l = 2

                           f           l = 3

For this problem l = 2

3.- Determine the value of m

when l = 2, m takes values of -2, - 1, 0, 1 or 2

3 0
3 years ago
How many total ions are in 2.95 g of magnesium sulfite?
Aleksandr [31]

Answer:

Approximately 5.65 \times 10^{-2}\; \rm mol, which is approximately 3.41 \times 10^{22} particles.

Explanation:

Formula of magnesium sulfite: \rm MgSO_3.

Look up the relative atomic mass of \rm Mg. \rm S, and \rm O on a modern periodic table:

  • \rm Mg: 24.305.
  • \rm S: 32.06.
  • \rm O: 15.999.

The ionic compound \rm MgSO_3 consist of magnesium ions \rm {Mg}^{2+} and sulfite ions \rm {SO_3}^{2-}.

Notice that \rm {Mg}^{2+}\! and \rm {SO_3}^{2-}\! combine at a one-to-one ratio to form the neural compound \rm MgSO_3. Therefore, each \rm MgSO_3\! formula unit would include one \rm {Mg}^{2+} ion and one \rm {SO_3}^{2-} ion (that would be two ions in total).

Calculate the formula mass of one such formula unit:

\begin{aligned}&\; M(\mathrm{MgSO_3}) \\ = & \; 24.305 + 32.06 + 3 \times 15.999 \\ = & \; 104.362\; \rm g \cdot mol^{-1}\end{aligned}.

In other words, the mass of one mole of \rm MgSO_3 formula units (which includes one mole of \rm {Mg}^{2+} ions and one mole of \rm {SO_3}^{2-} ions) would be 104.362\; \rm g.

Calculate the number of moles of such formula units in that 2.95\; \rm g of \rm MgSO_3:

\begin{aligned}n&= \frac{m}{M}\\ &=\frac{2.95\; \rm g}{104.362\; \rm g \cdot mol^{-1}} \approx 2.82670 \times 10^{-2}\; \rm mol\end{aligned}.

There are two moles of ions in each mole of \rm MgSO_3 formula units. Therefore, that 2.82670 \times 10^{-2}\; \rm mol of \rm MgSO_3\! formula units would include approximately 2 \times 2.82670 \times 10^{-2}\; \rm mol \approx 5.65\times 10^{-2}\; \rm mol of ions.

8 0
3 years ago
Select all that apply. All living things... <br><br> Can someone check this for me?
Maksim231197 [3]

your answers are right.

6 0
3 years ago
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