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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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You are given solutions of hcl and naoh and must determine their concentrations. you use 27.5 ml of naoh to titrate 100.0 ml of
Dafna1 [17]
1) Start by standardizing the solution of NaOH by using the solution of H2SO4 whose concentration is known.

2) Equation:

2Na OH + H2SO4 --> Na2 SO4 + 2H2O

3) molar ratios

2 mol NaOH : 1 mol H2SO4

4) Number of moles of H2SO4 in 50.0 ml of 0.0782 M solution

M = n / V => n = M*V = 0.0782 M * 0.050 l = 0.00391 mol H2SO4

5) Number of moles of NaOH

2 moles NaOH / 1mol H2SO4 * 0.00391 mol H2SO4 = 0.00782 mol NaOH

6) Concentration of the solution of NaOH

M = n / V = 0.00782 mol / 0.0184 ml = 0.425 M

7) Standardize the solution of HCl

Chemical reaction:

NaOH + HCl --> NaCl + H2O

8) Molar ratios

1 mol NaOH : 1 mol HCl

9) Number of moles of NaOH in 27.5 ml

M = n / V => n = M * V = 0.425 M * 0.0275 l = 0.01169 moles NaOH

10) Number of moles of HCl

1 mol HCl / 1mol NaOH * 0.01169 mol NaOH = 0.01169 mol HCl

11) Concentration of the solution of HCl

M = n / V = 0.01169 mol / 0.100 l = 0.1169 M

Rounded to 3 significant figures = 0.117 M

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3 0
3 years ago
Find the mass in grams of 3.00 x 1023 molecules of F2
LUCKY_DIMON [66]
<h3>Answer:</h3>

18.9 g F₂

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Reading a Periodic Table
  • Using Dimensional Analysis
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<h3>Explanation:</h3>

<u>Step 1: Define</u>

3.00 × 10²³ molecules F₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

Molar Mass of F₂ (Diatomic) - 38.00 g/mol

<u>Step 3: Convert</u>

  1. Set up:                              \displaystyle 3.00 \cdot 10^{23} \ molecules \ F_2(\frac{1 \ mol \ F_2}{6.022 \cdot 10^{23} \ molecules \ F_2})(\frac{38.00 \ g \ F_2}{1 \ mol F_2})
  2. Multiply:                                                                                                             \displaystyle 18.9306 \ g \ F_2

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

18.9306 g F₂ ≈ 18.9 g F₂

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