The standard International System of Units (SI) unit of mass is the kilogram(kg). The kilogram is 1000 grams (g), first defined in 1795 as one cubic decimeter of water at the melting point of ice.
Answer:
option C= hydrolysis and break down
Explanation:
All other three pairs are correct coupling of each others.
Option A= dehydration synthesis and hydrolysis
Dehydration synthesis:
In dehydration synthesis monomers combine through the covalent bonds and form large molecules. The large molecules are called polymers. The water as a byproduct also released when monomers joints together.
Hydrolysis:
In hydrolysis the polymers are break down into monomers by using water molecules. The catalysts are also required in this process.
Option B= Catabolic and Anabolic
Anabolic:
In this process smaller molecules combine to gather to form large complex molecules by using energy.
For example simple glucose molecules join together to form large disaccharides.
Catabolic:
It is the break down of large complex molecules to the smaller molecules.
For example during cellular respiration sugar molecules break down and generate energy.
Option D= Break down and synthesis
The break down and synthesis are also reverse pair of each others. The synthesis involve the formation of molecules form smaller component while the break down involve destruction of molecules into smaller units.
Answer:
4.1x10⁻⁵
Explanation:
The dissociation of an acid is a reversible reaction, and, because of that, it has an equilibrium constant, Ka. For a generic acid (HA), the dissociation happens by:
HA ⇄ H⁺ + A⁻
So, if x moles of the acid dissociates, x moles of H⁺ and x moles of A⁻ is formed. the percent of dissociation of the acid is:
% = (dissociated/total)*100%
4.4% = (x/[HA])*100%
But x = [A⁻], so:
[A⁻]/[HA] = 0.044
The pH of the acid can be calcualted by the Handersson-Halsebach equation:
pH = pKa + log[A⁻]/[HA]
3.03 = pKa + log 0.044
pKa = 3.03 - log 0.044
pKa = 4.39
pKa = -logKa
logKa = -pKa
Ka = 
Ka = 
Ka = 4.1x10⁻⁵
Answer:
An addiction could occur, maybe an overdose?, this could lead to death and maybe you would do unreasonable things which could get you fined or arrested.
Explanation:
Answer:
The concentration of cyclobutane after 875 seconds is approximately 0.000961 M
Explanation:
The initial concentration of cyclobutane, C₄H₈, [A₀] = 0.00150 M
The final concentration of cyclobutane, [
] = 0.00119 M
The time for the reaction, t = 455 seconds
Therefore, the Rate Law for the first order reaction is presented as follows;
![\text{ ln} \dfrac {[A_t]}{[A_0]} = \text {-k} \cdot t }](https://tex.z-dn.net/?f=%5Ctext%7B%20ln%7D%20%5Cdfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%20%3D%20%5Ctext%20%7B-k%7D%20%5Ccdot%20t%20%7D)
Therefore, we get;
![k = \dfrac{\text{ ln} \dfrac {[A_t]}{[A_0]}} {-t }](https://tex.z-dn.net/?f=k%20%3D%20%5Cdfrac%7B%5Ctext%7B%20ln%7D%20%5Cdfrac%20%7B%5BA_t%5D%7D%7B%5BA_0%5D%7D%7D%20%20%7B-t%20%7D)
Which gives;
k ≈ 5.088 × 10⁻⁴ s⁻¹
The concentration after 875 seconds is given as follows;
[] = [A₀]·
Therefore;
[] = 0.00150 × 
= 0.000961
The concentration of cyclobutane after 875 seconds, [] ≈ 0.000961 M
