Answer:
(a) 1:3
(b)
mmol of OH⁻ = 1.893 mmol
mmol of H₃C₆H₅O₇: 0.6310 mmol
mass of H₃C₆H₅O₇: 0.1212 g
mass % of H₃C₆H₅O₇ in orange juice: 1.37%
molarity of H₃C₆H₅O₇ in orange juice: 0.0726 M
Explanation:
Let's consider the following balanced chemical equation.
H₃C₆H₅O₇ + 3 OH⁻ → C₆H₅O₇³⁻ + 3 H₂O
<em>(a) What is the stoichiometry of H₃C₆H₅O₇ to OH⁻? </em>
The molar ratio of H₃C₆H₅O₇ to OH⁻ is 1:3
<em>(b)</em>
<em>mmol of OH⁻: ?</em>

<em>mmol of H₃C₆H₅O₇: ?</em>

<em>mass of H₃C₆H₅O₇: ?</em>
The molar mass of the citric acid is 192.1 g/mol.

<em>mass % of H₃C₆H₅O₇ in orange juice: ?</em>

<em>molarity of H₃C₆H₅O₇ in orange juice: ?</em>

Answer:
Electron-pair geometry: tetrahedral
Molecular geometry: trigonal pyramidal
Hybridization: sp³
sp³ - 4 p
Explanation:
There is some info missing. I think this is the original question.
<em>For NBr₃, What are its electron-pair and molecular geometries? What is the hybridization of the nitrogen atom? What orbitals on N and Br overlap to form bonds between these elements?</em>
<em>The N-Br bonds are formed by the overlap of the ___ hybrid orbitals on nitrogen with ___ orbitals on Br.</em>
<em />
Nitrogen is a central atom surrounded by 4 electron domains. According to VESPR, the corresponding electron-pair geometry is tetrahedral.
Of these 4 electron domains, 3 represent covalent bonds with Br and 1 lone pair. According to VESPR, the corresponding molecular geometry is trigonal pyramidal.
In the nitrogen atom, 1 s orbital and 3 p orbitals hybridize to form 4 sp³ orbitals for each of the electron domains.
The N-Br bonds are formed by the overlap of the sp³ hybrid orbitals on nitrogen with 4p orbitals on Br.
In glycolysis, glucose molecule is converted into pyruvate molecules .
It can't be cellular respiration because it starts always with glucose i.e glycolysis. But here it is the process after glycolysis (starting from pyruvate) and changing into lactic acid. It's an aerobic respiration called as lactic acid fermentation.
Fermentation
Answer:
Is there another picture with the #'s in the boxes
Explanation: