Answer:
The factors that govern the position of an IR absorption peak are:
(A) strength of the bond
(C) masses of the atoms involved in the bond
(D) the type of vibration being observed
Explanation:
In infrared spectroscopy the molecules absorb the frequencies that are characteristic of their structure. These absorptions occur at resonance frequencies, that is, the frequency of the absorbed radiation coincides with the frequency of vibration. The energies are affected by the shape of molecular potential energy surfaces, the masses of atoms and the associated vibronic coupling. The resonance frequencies are also related to the strength of the bond and the mass of the atoms at each end of it. Therefore, the frequency of vibrations is associated with a particular normal movement mode and a particular type of link.
Answer:
Esterification reaction
Explanation:
When we have to go from an acid to an ester we can use the esterification reaction. On this reaction, an alcohol reacts with a carboxylic acid on acid medium to produce an ester and water. (See figure).
In this case, we need the methyl ester, therefore we have to choose the appropriate alcohol, so we have to use the methanol as reactive if we have to produce the methyl ester.
a. 2-keto acid decarboxylase AND alcohol dehydrogenase
b. alcohol dehydrogenase
c. transaminase
d. acetolactate synthasee. 2-ketoacid decarboxylase
Answer:
A. 2-keto acid decarboxylase and alcohol dehydrogenase
Explanation:
2-keto acid decarboxylase and alcohol dehydrogenase are used to produce many higher alcohols. These enzymes also display a high degree of specificity on their substrate
Answer:
-68.4 kJ
Explanation:
The standard enthalpy of vaporization = 23.3 kJ/mol
which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).
To calculate heat released when 50.0 g of ammonia is condensed at -33 °C.
This is the opposite of enthalpy of vaporization which means that same magnitude of heat is released.
Thus, Q = -23.3 kJ/mol
Where negative sign signifies release of heat
Given: mass of 50.0 g
Molar mass of ammonia = 17.034 g/mol
Moles of ammonia = 50.0 /17.034 moles = 2.9353 moles
Also,
1 mole of ammonia when condenses at -33 °C releases 23.3 kJ
2.9412 moles of ammonia when condenses at -33 °C releases 23.3×2.9353 kJ
Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.
The heat released when 50.0 g of ammonia condenses at its boiling point is -68.4 kJ. This is calculated by multiplying the moles of ammonia by the enthalpy of vaporization and recognizing that heat is released in condensation.
To solve this problem, we need to understand the concept of enthalpy of vaporization, which is the heat needed to convert 1 mole of a substance from a liquid to a gas at constant pressure and temperature. For ammonia (NH3), which boils at -33 °C, the enthalpy of vaporization is 23.3 kJ/mol. However, we want the heat released when 50.0 g (around 2.94 moles) of ammonia condenses, which is the reverse process of vaporization. Thus, the energy would be released rather than absorbed.
Now, let's calculate this value. We multiply the number of moles of ammonia by the enthalpy of vaporization:
2.94 moles x 23.3 kJ/mol = 68.4 kJ
Since this is the reverse of the process of vaporization, heat is released, so the enthalpy change is negative (-68.4 kJ). Therefore, the correct answer is -68.4 kJ.
#SPJ11
Answer:
0.0192 mL per min.
Explanation:
IV rate = 36 mg per 30 min.
IV concentration = 125 mg per 2.0 mL
36 mg per 30 min. IV rate = 36/30 = 1.2 mg per min
If 125 mg methylprednisolone is present in 2.0 mL of the IV nag, how many mL would contain 1.2 mg?
= 2x1.2/125
= 0.0192 mL
Therefore, the flow rate of the IV must be 0.0192 mL per min. in order to be able to deliver 36 mg per 30 min.
Answer:
Molarity= 4M
Explanation:
n= CV
24= C×6,
C= 24/6 = 4M
Answer:4M
Explanation:
Number of moles=24
Volume=6L
Molarity=number of moles ➗ volume
Molarity=24 ➗ 6
Molarity=4M
Answer:
Is it Group?
Explanation: