Answer:
The rocket is now too heavy to reach its destination.
So you wont die from eating too much great food! XD
The m/z and relative abundance of the ions contributed to the peak at 21.876 min. The relative abundance will be 21.876%.
<h3>
What is relative abundance?</h3>
- The proportion of atoms with a particular atomic mass present in an element sample taken from a naturally occurring sample is known as the relative abundance of an isotope.
- When the relative abundances of an element's isotopes are multiplied by their atomic masses and the results are added up, the result is the element's average atomic mass, which is a weighted average.
- Chemists often divide the number of atoms in a particular isotope by the sum of the atoms in all the isotopes of that element, then multiply the result by 100 to determine the percent abundance of each isotope in a sample of that element.
To learn more about relative abundance with the given link
brainly.com/question/1594226
#SPJ4
"The uncertainty<span> in </span>velocity<span> is Δv=1.05⋅105m/s . According to the Heisenberg </span>Uncertainty<span> Principle, you cannot measure simultaneously with great precision both the momentum and the position of a particle. m - the mass of an electron - 9.10938⋅10−31kg."
-socratic.com</span>
Answer: The actual free-energy change for the reaction -8.64 kJ/mol.
Explanation:
The given reaction is as follows.
Fructose 1,6-bisphosphate 
Glyceraldehyde 3-phosphate + DHAP
For the given reaction, 
is 23.8 kJ/mol.
As we know that,
Here, R = 8.314 J/mol K, T = 
= (37 + 273) K
= 310.15 K
Fructose 1,6-bisphosphate = 
M
Glyceraldehyde 3-phosphate = 
M
DHAP = 
M
Expression for reaction quotient of this reaction is as follows.
Reaction quotient = ![\frac{[DHAP][\text{glyceraldehyde 3-phosphate}]}{[/text{Fructose 1,6-bisphosphate}]}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BDHAP%5D%5B%5Ctext%7Bglyceraldehyde%203-phosphate%7D%5D%7D%7B%5B%2Ftext%7BFructose%201%2C6-bisphosphate%7D%5D%7D)
Q = 
= 
Now, we will calculate the value of 
as follows.
= 
= -8647.73 J/mol
= -8.64 kJ/mol
Thus, we can conclude that the actual free-energy change for the reaction -8.64 kJ/mol.
