What can we consider to be a strong base and a strong acid?
Well there are 7 strong acids that I recommend to remember. To be considered a strong acid, the acid has to completely dissociate or ionize in a solution, which means that a H+ is completely gone from the acid. These are the acids: 1. HCl (Hydrochloric acid) 2. HBr (Hydrobromic acid) 3. HI (Hydroiodic acid) 4. HNO3 (Nitric acid) 5. HClO3 (Chloric acid) 6. HClO4 (Perchloric acid) 7. H2SO4 (Sulfuric acid) (Only first ionization) Likewise, strong bases are compounds that will break completely to form OH- groups in a solution. These are considered strong bases: 1. Metals from group 1A + OH a. LiOH (Lithium hydroxide) b. NaOH (Sodium hydroxide) c. KOH (Potassium hydroxide) d. RbOH (Rubidium hydroxide) e. CsOH (Cesium hydroxide) 2. Heavy metals from group 2A + OH a. Ca(OH)2 (Calcium hydroxide) b. Sr(OH)2 (Strontium hydroxide) c. Ba(OH)2 (Barium hydroxide) These are the acids and bases that you will mostly use during general chemistry classes! You will approach this concept later in other chemistry classes as well.
Let's say that you have the acceleration function of a certain particle trajectory. And you are asked to find the position function of the particle. How are you able to find the position function only from the acceleration function?
This is a topic that is related to integration. From the derivative concepts, we know that if you find the 1st derivative of a position function you will have the velocity function. And if you find 2nd derivative position function or the 1st derivative of the velocity function from before you will have the acceleration function. Now the concept of integration you could see it as an anti-derivation, which means that you will go backwards in the process of finding the derivative of a function. That means that if you have the acceleration function and you want to find the position function you will have to follow these steps: 1. Integrate the acceleration function. 2. After integrating the acceleration function, you will obtained the velocity function. 3. If you integrate the velocity function then, you will obtained the position function And now you could track specific positions points of the particle during a certain interval.
Let's say that you have a parabola and I want to know the solutions to the x interceptions. What would you use to find the roots of the parabola function?
You can use the quadratic formula! If you have a parabola function that has the form of ax^2 + bx + c = 0, you can applied the the quadratic formula. The quadratic formula will give you 2 roots solutions: 1. x = [ -b + sqrt( b^2 - 4ac) ] / (2b) 2. x = [ -b - sqrt( b^2 - 4ac) ] / (2b) This is a very important equation that has many applications in later courses such as Calculus, Pre-Calculus, Linear Algebra, among others!