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timofeeve [1]
2 years ago
7

The hormone epinephrine is released in the human body during stress and increases the body's metabolic rate. Epinephrine, like m

any biochemical compounds, is composed of carbon, hydrogen, oxygen, and nitrogen. The percentage composition of the hormone is 59.0% C, 7.15% H, 26.2% O, and 7.65% N. Determine the empirical formula.
Chemistry
1 answer:
mart [117]2 years ago
5 0

Answer:

C9 H13 N O3  

Explanation:

Carbon mole wt = 12

   59 /12 = 4.917

Hydrogen  mole wt = 1

  7.15 / 1 =   7.15

Oxygen mole wt = 16

  26.2 / 16 = 1.6375

Nitrogen  mole wt = 14

  7.65 / 14 = .5464                  

Now divide the numbers by the smallest found      

 Carbon   4.917 / .5464 = 9

 Hydrogen  7.15 / .5464 = 13

 Oxygen    1.6375 / 3

  Nitrogen  .5464 /.5464 =1                   emp form  C9 H13 O3 N      

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Answer:d

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If the formula for potassium chlorate is KClO3 and the formula for magnesium fluoride is MgF2, then what isthe formula for magne
mr_godi [17]
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5 0
3 years ago
Calculate the enthalpy of the reaction below (∆Hrxn, in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g).
nalin [4]

The enthalpy change of the reaction below (ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

The bond energies data is given as follows:

BE  for C≡O  = 1072 kJ/mol

BE for Cl-Cl = 242 kJ/mol

BE for C-Cl = 328 kJ/mol

BE for C=O = 766 kJ/mol

The enthalpy change for the reaction is given as :

ΔHr×n = ∑H reactant bond - ∑H product bond

ΔHr×n = ( BE C≡O + BE Cl-Cl) - ( BE C=O + BE 2 × Cl-Cl )

ΔHr×n = ( 1072 + 242 ) - ( 766 + 656 )

ΔHr×n = 1314 - 1422

ΔHr×n = - 108 kJ

Thus, The enthalpy change of the reaction below ( ΔHr×n , in kJ) using the bond energies provided. CO(g) + Cl₂(g) → Cl₂CO(g). is - 108kJ.

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7 0
1 year ago
5)
Kryger [21]

The empirical formula : C₁₂H₄F₇

The molecular formula : C₂₄H₈F₁₄

<h3>Further explanation</h3>

mol C (MW=12 g/mol)

\tt \dfrac{18.24}{12}=1.52

mol H(MW=1 g/mol) :

\tt \dfrac{0.51}{1}=0.51

mol F(MW=19 g/mol)

\tt \dfrac{16.91}{19}=0.89

mol ratio of C : H : O =1.52 : 0.51 : 0.89=3 : 1 : 1.75=12 : 4 : 7

Empirical formula : C₁₂H₄F₇

(Empirical formula)n=molecular formula

( C₁₂H₄F₇)n=562 g/mol

(12.12+4.1+7.19)n=562

(281)n=562⇒ n =2

Molecular formula : C₂₄H₈F₁₄

6 0
3 years ago
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