Every Topic is equally easy when concepts gets cleared.

@@Edusetucbsespeedmaths but school system doesn't understand that :/

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to understand this we must understand ionic bonds and ionic solids! this will all be revealed as you move forward, in the tutorials about the chemical bond, intermolecular forces, solutions, etc.

The reason F falls off of Ag and becomes part of Ca is because F is more attracted to Ca than Ag. The Ca, which is Calcium, has two electrons that it gives to F. The only problem is, F would love to accept one electron from Ca but not two. The solution? F goes and gets another F to join him and the two F's each keep one electron from Ca. They now become a team of three: One Ca and two F's. This is what happens in chemical reactions, they change partners.

@@ProfessorDaveExplains how can we do the comprehension no. 4?

@@johnmayes9649 they sound like swingers. Haha

check out my tutorial on balancing redox reactions

@@ProfessorDaveExplains thank u so much Professor 🖤

yes i think just repetition! you start to get an intuition for which element to balance first and what not. also check for polyatomic ions, those can be balanced as a whole, without having to split them up into elements.

I recommend up setting up a system of linear equations to do it. Assign a lowercase letter as a placeholder for each coefficient, and go in alphabetical order for each molecule you have. Tally up the number of atoms for each elemental species, in terms of your coefficients. Assume the first coefficient a=1. Then solve the system of linear equations, just like it were a math problem. Matrices can help when you have numerous equations. It is likely that you will get fractional coefficients, and when you do, this just means you need to multiply a by the least common multiple of all the denominators, so that every coefficient becomes an integer. For instance, given the burning of Isopropyl alcohol: C3H8O + O2 -> CO2 + H2O Assign lowercase letters as coefficients: a C3H8O + b O2 -> c CO2 + d H2O Equate counts of each elemental species on each side: Carbon: 3*a = c Hydrogen: 8*a = 2*d Oxygen: a + 2*b = 2*c + d Assume a = 1 Solve for our unknowns: a=1 b=9/2 c=3 d=4 Because we got a fractional result for b, i.e. 9/2, we have to multiply all of the above numbers by 2, to make them all whole numbers. Our balanced equation becomes: 2 C3H8O + 2 O2 -> 6 CO2 + 8 H2O

Nevermind...... i realized i was looking at hydrogen not oxygen haha

The triple bond of nitrogen is a lot more energy intensive to break, that most of the nitrogen gas in the air is just a spectator gas to the fire. There is about as much nitrogen in the exhaust as there is in the original air. Some of it is consumed to make unintended products of combustion like NO and NO2, but most of it remains in its original compound of elemental nitrogen N2. The nitrogen gas might as well be an inert gas, for most chemical reactions.

not sure what you mean!

Did you get it? Hope you did but I am showing you anyways(there is no trick btw) H3PO4 + KOH ----> K3PO4 + H2O H - 4 | H - 2 P - 1 | P - 1 O - 5 | O - 5 K - 1 | K - 3 ( have this in your mind or jot it down , you don't have to correct anything in the main equation yet, first we are gonna figure out the numbers to be written in the eqn.) 1. H first. Not balanced. 2. Then P . Its 1 on both sides. Balanced. 3. O. Okay.... 4. Finally K . It is 3 on the product side. We have to make the reactant side also 3. There is only one way. We write 3 on the reactant side. ( KOH changes to 3KOH) Current eqn : H3PO4 + 3KOH -----> K3PO4 + 3H2O 5. Okay..now return to H. (See the current eqn)We've got total six H in the reactant side( Three H of H3PO4 and 3KOH) And also six H in the product side..... 3H2O (3×2) H balanced:)) 6. Now we have O. There are seven O in the reactant side ( four in H3PO4 and three in 3KOH .. 4+3). And also seven in product side.( four in K3PO4 and 3 in 3H2O....4+3) O is balanced. [ Just like that . You don't have to write the figures at first, estimate them and keep looking ] Therefore balanced eqn:)) #4)H3PO4 + KOH ----> K3PO4 + H2O H3PO4 + 3KOH -----> K3PO4 + 3H2O

@@devikagopakumar8906 Thank you man

@@-_wkwyy You're welcome bro...

not anymore, focusing on science communication!

I recommend a linear algebra approach. Assign a variable as the coefficient of each molecule. Go in alphabetical order, and use lowercase letters. Set up a system of linear equations to equate the population of each element's atoms on both sides of the equation, and solve by your algebraic method of choice. Assume the first coefficient a=1, so that your number of equations matches your number of unknowns. From time to time, you will get fractional coefficients with the above method. This is not acceptable as an answer in Chemistry, because you can't react a fraction of a molecule. Therefore, you then have to multiply through to clear all the fraction denominators. Find the least common multiple of all the fraction denominators, and multiply through by that number. You will then have the smallest possible group of whole number coefficients that balance the chemical equation.

"He knows a lot about the science stuff, Professor Dave explains!"

@@ProfessorDaveExplains he realy does reply

@@ProfessorDaveExplains that means he is a true teacher

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95,587 hailed from the Mesopotamia lands

You cannot create or destroy atoms, by ordinary chemical means. You also cannot change the elemental species of any given atom. What we do is tally up each element's atom on each side of the chemical equation, and apply coefficients to the front of each chemical formula, indicating that we need at least that many of that molecule, in order to get a full reaction. Consider the burning of octane gasoline in oxygen. C8H18 + O2 => CO2 + H2O We apply placeholder letters as coefficients to each formula: a C8H18 + b O2 => c CO2 + d H2O Set up a system of equations to solve for a, b, c, and d. Assume a to equal 1. We'll scale up the coefficients if we get fractional values. Carbon: 8*a = c Hydrogen: 18*a = 2*d Oxygen: 2*b = 2*c + d Set a=1, and solve for b, c, and d: b=12.5 c=8 d=9 Since we end up with a fractional value for b, this means we need to set a=2, and double each of the remaining coefficients. All the coefficients need to end up as a whole number, so it has a real life significance, since you can't react half of a molecule. a = 2 b = 25 c = 16 d = 18 This means our balanced equation is: 2 C8H18 + 25 O2 => 16 CO2 + 18 H2O In English, this would be read aloud as: "Two molecules of octane react with 25 molecules of oxygen gas. The reaction yields 16 molecules of carbon dioxide and 18 molecules of water vapor."

@@carultch thanks it is s much clearer now.

Okay scratch that I am familiar with it. But still better understanding is better regardless

Uhm is this a joke or your actually wondering. I'm uncertain because of your profile picture.

I'm assuming you should already know because of your profile picture, but to still answer the question : no. I wouldn't call NaCl a molecule.

@@setonix9151 sodium chloride is not a molecule, It is an ionic crystal. We write it as NaCl because they bond together 1:1. I am pretty sure that this video is wrong about that.

@@setonix9151 it is a joke lol

@@grandiflorum4716 i did just say I wouldn't call it a molecule Edit : and I used NaCl because I'm lazy

Yes I keep replaying too, and I’m still confused

You don't necessarily care where one particular atom goes. You simply need to generate coefficients on each chemical formula, in order to have an equal number of each elemental species of each atom, on both sides. This is because you cannot create or destroy atoms, or change the elemental species of an atom, by ordinary chemical means. The way I recommend balancing equations: 1. Assign a lowercase letter on each molecule in the equation, as a placeholder for the coefficient. Go in alphabetical order, from a being assigned as the coefficient on the first formula within the equation. 2. Set up an equation from each known subscript on each atom in each compound, equating the total population of each kind of atom, across the reaction arrow. 3. You generally will have 1 fewer equation than you have unknowns, which means you'll have to assume one of your coefficients to arbitrarily equal 1. Assume a=1, and solve the system of linear equations. 4. If you get fractional answers, that means you just need to multiply all your solutions by a number that clears all the denominators. For instance, if you get a=1, b=1/2, c=3, and d=5 as your coefficients, just multiply each one of them by 2, to clear the fraction on b. This would make a=2, b=1, c=6, and d=10. I just made up this example.