Answer:
13,2 %.
Explanation:
Overall yield = 0.55 * 0.24
= 0.132
= 13,2 %.
Answer and Explanation:
1. Arrhenius Theory which describes the concept protonic. The substance that gives H+ ions when diluted in water is called as an acid (e.g. HCl) and the substance that dissociates OH-ions whenever it is diluted in water is called as the base (e.g. NaOH)
on the other hand
Bronsted Lowery Theory describes the concept of a proton donor-acceptor. The proton-donating species is an acid and the proton-accepting species is known as a base.
2. The Chemical name and nature of acid is shown below:-
Nature Chemical Name
a. HCl Acidic Hydrochloric Acid
b. KOH Basic Potassium hydroxide
c. HNO Acidic Nitric Acid
d. Mg(OH)2 Basic Magnesium hydroxide
Answer:
A. Arginine
Explanation:
The urea cycle is the cycle of the biochemical reactions which produces urea from ammonia.
Steps of the urea cycle:
- Carbamoyl phosphate, in presence of ornithine transcarbamoylase is converted to citrulline by the denotation of carbamoyl phosphate groupto ornithine and a phosphate group is released.
- Amino group of the aspartate and carbonyl group of the citrulline are condensed to form argininosuccinate in the presence of enzyme, argininosuccinate synthetase. This condensation reaction is ATP dependent.
- <u>Argininosuccinate then undergoes cleavage by the argininosuccinase to form intermediate, arginine and fumarate.</u>
- <u>Arginine is then cleaved by the arginase to form urea as well as ornithine.</u> Ornithine is transported back to mitochondria to begin urea cycle again.
Answer:
-10778.95 J heat must be removed in order to form the ice at 15 °C.
Explanation:
Given data:
mass of steam = 25 g
Initial temperature = 118 °C
Final temperature = 15 °C
Heat released = ?
Solution:
Formula:
q = m . c . ΔT
we know that specific heat of water is 4.186 J/g.°C
ΔT = final temperature - initial temperature
ΔT = 15 °C - 118 °C
ΔT = -103 °C
now we will put the values in formula
q = m . c . ΔT
q = 25 g × 4.186 J/g.°C × -103 °C
q = -10778.95 J
so, -10778.95 J heat must be removed in order to form the ice at 15 °C.
The question provides the data in an incorrect way, but what the question is asking is for the entropy change when combining 3 moles of water at 0 °C (273.15 K) with 1 mole of water at 100 °C (373.15 K). We are told the molar heat capacity is 75.3 J/Kmol. We will be using the following formula to calculate the entropy change of each portion of water:
ΔS = nCln(T₂/T₁)
n = number of moles
C = molar heat capacity
T₂ = final temperature
T₁ = initial temperature
We can first find the equilibrium temperature of the mixture which will be the value of T₂ in each case:
[(3 moles)(273.15 K) + (1 mole)(373.15 K)]/(4 moles) = 298.15 K
Now we can find the change in entropy for the 3 moles of water heating up to 298.15 K and the 1 mole of water cooling down to 298.15 K:
ΔS = (3 moles)(75.3 J/Kmol)ln(298.15/273.15)
ΔS = 19.8 J/K
ΔS = (1 mole)(75.3 J/Kmol)ln(298.15/373.15)
ΔS = -16.9 J/K
Now we combine the entropy change of each portion of water to get the total entropy change for the system:
ΔS = 19.8 J/K + (-16.9 J/K)
ΔS = 2.9 J/K
The entropy change for combining the two temperatures of water is 2.9 J/K.
