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nekit [7.7K]
3 years ago
9

The reaction converting glycerol to glycerol-3-phosphate (energetically unfavorable) can be coupled with the conversion of ATP t

o ADP (energetically favorable): glycerol+HPO42−ATP+H2O⟶⟶glycerol-3-phosphate+H2OADP+HPO42−+H+ Part A What are the net products of the coupled reactions above?
Chemistry
1 answer:
Brut [27]3 years ago
8 0

Answer:

glycerol-3-phosphate, ADP, H⁺

Explanation:

The reaction of converting glycerol to glycerol-3-phosphate which makes is unfavorable and is coupled with the second reaction which involves conversion of ATP to ADP which is high energetically favorable.

Reaction 1:  Glycerol + HPO₄²⁻ ⇒ Glycerol-3-phosphate + water

Reaction 2:             ATP  + H₂O ⇒ ADP + HPO₄²⁻ + H⁺

The coupled reaction of both the reactions become favorable. Thus, the overall coupled reaction is:

<u>Glycerol + ATP ⇒ Glycerol-3-phosphate + ADP + H⁺</u>

The net products are = glycerol-3-phosphate, ADP, H⁺

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When FeC13 is ignited in an atmosphere of pure oxygen, this reaction takes place. 4FeCl3(sJ 30lgJ ~ 2F~0 (sJ 6Cl2(gJ If 3.00 mol
oksano4ka [1.4K]

Answer : The reagent present in excess and remains unreacted is, O_2

Solution : Given,

Moles of FeCl_3 = 3.00 mole

Moles of O_2 = 2.00 mole

Excess reagent : It is defined as the reactants not completely used up in the reaction.

Limiting reagent : It is defined as the reactants completely used up in the reaction.

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

2FeCl_3(s)+O_2(g)\rightarrow 2FeO(s)+3Cl_2(g)

From the balanced reaction we conclude that

As, 2 moles of FeCl_3 react with 1 mole of O_2

So, 3.00 moles of FeCl_3 react with \frac{3.00}{2}=1.5 moles of O_2

From this we conclude that, O_2 is an excess reagent because the given moles are greater than the required moles and FeCl_3 is a limiting reagent and it limits the formation of product.

Hence, the reagent present in excess and remains unreacted is, O_2

4 0
3 years ago
Find the total number of atoms in a sample of cocaine hydrochloride, c17h22clno4, of mass 23.0 mg .
Nataliya [291]
From the periodic table:
mass of carbon = 12 grams
mass of hydrogen = 1 gram
mass of nitrogen = 14 grams
mass of oxygen = 16 grams
mass of chlorine = 35.5 grams
Therefore,
molar mass of <span>c17h22clno4 = 17(12) + 22(1) + 35.5 + 14 + 4(16) = 339.5 grams

number of moles = mass / molar mass
number of moles = (23*10^-3) / (339.5)
number of moles = 6.77 * 10^-5 moles

number of atoms = number of moles * Avogadro's number
number of atoms = 6.77*10^-5 * 6.022*10^-23
number of atoms = 4.079 * 10^-27 atoms</span>
3 0
4 years ago
1. Complete the following
Marat540 [252]

Answer:

\large \boxed{\text{0.603 mol}}

Explanation:

We must do the conversions

mass of C₆H₁₂O₆ ⟶ moles of C₆H₁₂O₆ ⟶ moles of O₂

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:        180.16

           C₆H₁₂O₆ + 6O₂ ⟶ 6CO₂ + 6H₂O

m/g:        18.1

(a) Moles of C₆H₁₂O₆

\text{Moles of C$_{6}$H$_{12}$O}_{6} = \text{18.1 g C$_{6}$H$_{12}$O}_{6}\times \dfrac{\text{1 mol C$_{6}$H$_{12}$O}_{6}}{\text{180.16 g C$_{6}$H$_{12}$O}_{6}}\\\\= \text{0.1005 mol C$_{6}$H$_{12}$O}_{6}

b) Moles of O₂

\text{Moles of O}_{2} =\text{0.1005 mol C$_{6}$H$_{12}$O}_{6} \times \dfrac{\text{6 mol O}_{2}}{\text{1 mol C$_{6}$H$_{12}$O}_{6}} = \text{0.603 mol O}_{2}\\\\\text{The reaction requires $\large \boxed{\textbf{0.603 mol}}$ of oxygen}  

3 0
3 years ago
39. Analyze What subscripts would you most likely use if
Igoryamba

Based on their valencies, the subscripts of the ionic compounds formed will be:

  • 1 and 1
  • 2 and 1
  • 1 and 2
  • 2 and 2

<h3>What subscripts would you most likely use if the following substances formed an ionic compound?</h3>

A. An alkali metal and a halogen.

An alkali metal and a halogen both have valencies of one.

Therefore the subscripts would be 1.

B. An alkali metal and a non-metal from group 16.

An alkali metal has a valency of 1 and a group 16 non-metal has a valency of 2.

By exchange of valencies, the subscript would be 2 and 1.

C. An alkaline earth metal and a halogen.

An alkaline earth metal has a valency of 2 and a halogen has a valency of 1.

By exchange of valencies, the subscripts would be 1 and 2.

D. An alkaline earth metal and a non-metal from group 16.

An alkaline earth metal has a valency of 2 and a non-metal from group 16 has a valency of 2.

By exchange of valencies, the subscripts would be 2 and 2.

Therefore, the subscripts of the ionic compounds formed will be:

  • 1 and 1
  • 2 and 1
  • 1 and 2
  • 2 and 2

Learn more about about valency at: brainly.com/question/2284519

4 0
3 years ago
An air compressor takes atmospheric air (think about what pressure it is at) and pressures it up to 68 (psig). In the process, i
Aleks [24]

Answer:

5.19 m³  

Explanation:

Data Given:

initial Pressure P1 = atmospheric pressure

Reported atmospheric pressure =  14.696 psi

Final pressures P2 =  68 psi

initial Temperature T1 =  26 °C

final Temperature T2 =  48 °C

initial Volume V1= 13 m³

final Volume V2 = ?

Solution:

Formula will be used

                    P1 V1 / T1 = P2 V2 / T2

To calculate volume rearrange the above formula

                   V2 = (P1 V1) T2 / T1 P2 . . . . . . . .(1)

put values in equation 1

                  V2 = (14.696 psi x 13 m³) 48 °C / 26 °C x 68 psi

                  V2 = (191.05 psi m³) 48 °C / 1768 °C psi

                  V2 = (9170.3  psi m³ °C / 1768 °C psi

                  V2 = 5.19 m³

So, final volume is 5.19 m³                

7 0
4 years ago
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