Answer:
When C1 is labeled in glucose, it ends up in the methyl group of pyruvate.
Aldolase cleaves a hexose into two trioses.
[See the image attached].
Asterisk indicates the label.
When C1 is labeled in glucose, it ends up in the carboxyl group of pyruvate.
Explanation:
Polarity is defined as the development of partial charges on the atoms of a molecule. In a water molecule, there are hydrogen and oxygen atoms.
Due to the difference in electronegativity of both hydrogen and oxygen atom there is development of partial positive charge on hydrogen atom and a partial negative charge on oxygen atom.
So, when bond between hydrogen and oxygen will break down then it will form hydrogen ions (
) and oxygen ions (
).
Ion-dipole interactions are defined as the interactions that occur when an ion interacts with the dipole of a molecule.
When an electron is added to a neutral atom to convert it into a negative ion then the amount of change taking place in its energy is known as electron affinity.
So, oxygen atom has an affinity towards cations and hydrogen atom has an affinity for anions.
Thus, we can conclude that following interactions and processes contribute to the dissolution of ionic compounds in water:
1. Affinity of oxygen towards cations
2. Ion–dipole interactions
4. Hydration
6. Affinity of hydrogen towards anions
Answer: NH3
Explanation: N2 + H2 yields NH3.
Answer:
1.78 atm
Explanation:
From the question given above, the following data were obtained:
Mass of Cl = 355 g
Molecular weight of Cl = 70.9 g/mol
Volume (V) = 70.0 L
Temperature (T) = 30 °C
Pressure (P) =?
Next, we shall determine the number of mole in 355 g of chlorine (Cl). This can be obtained as follow:
Mass of Cl = 355 g
Molecular weight of Cl = 70.9 g/mol
Mole of Cl =.?
Mole = mass /molecular weight
Mole of Cl = 355 /70.9
Mole of Cl = 5 moles
Next, we shall convert 30 °C to Kelvin temperature. This can be obtained as follow:
T(K) = T(°C) + 273
Temperature (T) = 30 °C
Temperature (T) = 30 °C + 273 = 303 K
Finally, we shall determine the pressure of the gas as illustrated below:
Volume (V) = 70.0 L
Number of mole (n) = 5 moles
Temperature (T) = 303 K
Gas constant (R) = 0.0821 atm.L/Kmol
Pressure (P) =?
PV = nRT
P × 70 = 5 × 0.0821 × 303
P × 70 = 124.3815
Divide both side by 70
P = 124.3815 / 70
P = 1.78 atm
Therefore, the pressure of the gas is 1.78 atm