I'm studying for the SAT and each time I face a problem I come and see your videos . You approximately solved every type of question that I need to survive SAT . Thank you so much
That formula saved my life! Nowhere does my book mention it, and I've been scratching my head for a few hours now trying to figure out what was happening. Thank you so much!
Professor Organic Chemistry Tutor, thank you for a solid introduction to Mixture Problems in AP/ General Chemistry. Everything in life uses some form of Chemistry. Mixtures problems are common in all levels of Chemistry. The equation that's given in this video is a good starting point when solving mixture problems. This is an error free video/lecture on RU-vid TV with the Organic Chemistry Tutor.
Im in an Hvac class at my community college but have to take an applied math class. I have been struggling with this all week and you just made my life 100% easier god bless you so much.
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I thought this video help me to solve the problem in 10sec cause i can solve it for less than minute For the first problem. Just use 2 equation 2 unknown. x+y=20 y=20-x. Eq. 1 0.2(x) + 0.4(y) = 0.28(20) eq. 2 Put eq. 1 in eq. 2 0.2(x) + 0.4(20-x) = 0.28(20) x=12
Amazing video, i used to be really scared of mixture problems, you explained it amazingly, they never mention this formula in my book, nonetheless, worked 10outta10 times. thanks a lot!
6:37 problem 2. What is the concentration of the mixture? The mixture is 200 L (50 from the 15% alcohol solution and 150 from the 35% alcohol solution). let y = percentage of alcohol concentration of the mixture, then y 200 L = the total amount of alcohol in the mixture. The first solution has 7.5 L of alcohol (50 x 15%) and the second solution has 52.5 L (15% x 150) and this equal to total alcohol in the mixture of y 200 L. solving for y will give the concentration of alcohol of the mixture, so 7.5 + 52.5 = y 200 or 60=y 200 or y = 6/20 or 3/10, 3/10 is the same as 30/100 or 30%. Therefore the mixture has a 30% alcohol solution. Ans 30% 10:03 alcohol solution or 60 L of alcohol is in it.
thank you so much!!! i was literally about to cry cuz i didn't understand my math and i'm scared of my teacher cause he always wants to check how i'm doing in my math class, but now i know what's going on and i'm reliefed. thank you again, and have a great day
Thank you!! I took organic chemistry 20 years ago and was trying to calculate mixing solutions of 3% and 10% to get a 6% 3.5 ML solution, I need 2 ml of 3% and 1.5 ml of 10% to get my goal 6%. Thanks again :)
Thank you very much sir. Your lecture helps a lot for understanding the problems. I love how you do it, step by step sir. God bless and keep safe. 🙏🙏🙏🙏🙏😇😇😇😇😇😇
This works for 10:53 as well. 60L is 3* bigger than 20L. Adding 60L of water to a 20L 28% concentration can be represented as (0*3+28)/4 = 7% concentration
For the first problem. Just use 2 equation 2 unknown. x+y=20 y=20-x. Eq. 1 0.2(x) + 0.4(y) = 0.28(20) eq. 2 Put eq. 1 in eq. 2 0.2(x) + 0.4(20-x) = 0.28(20) x=12
Mass is conserved. Volumes of liquids are not necessarily. (10 L of A + 10 L of B will not necessarily produce 20 L of a mixture.) Differences can be significant. Without density data, isn't it better to use weights rather than volumes?
Hi, I get how to use the equation to solve these types of problems. However, I’m not too sure on how you came up with the equation. Could you explain how to derive the equation?
Wait…what? If the student mixed 40% and 20% and ended up with 20ml of 28% Are you saying adding 12 ml of the 20% gives him a 20 ml 32% ? I’m confused….Did he start with a 20 ml container of 28% or did he start with a container of 20 ml of 32%?
Will you please solve this problem? Brine is a solution of salt and water. If a tub contains 50 gallons of a 5% solution of brine, how much water must evaporate to change it to an 8% brine solution?
Geez that formula is SO HELPFUL!!!! BTW- just curious how come the % isn't converted to a decimal before multiplying? (I feel like I would do that systematically, and hinder the equation....)
no because the factor was cuz of the increase of the second L to the total L. You have to consider than it is only one compound. Water is not included in the equation so it doesn't affect it. Try to do the same to other compounds. He said that it is only done when adding water to a pure, only.
Yes, but I kept 28% as 28 for two reasons. (1) It will not affect my final answer. (2) My final answer will be a percentage and not decimal. You can do it both ways. You should still get the right answer.
problem 1 let p = liters of the 20% acidic solution, then that solution has 1/5 p liters of acid Since the total liters produced = 20 liters, then the other 40% acidic solution is (20-p) liters and thus has 2/5 (20-p) liters of acid. Both 20% and 40% acidic solution are said to produce 20 liters of a 28% acidic solution or (20 liters x .28) 5 and 3/5 liters of acid. equation , 1/5 p + 2/5 (20-p)= 5 3/5 or 1/5 p + 8- 2/5 p = 5 3/5 or -1/5 p= - 2 2/5 or 1/5 p= 2 2/5 or 1/5 p =12/5 multiply both sides by 5 gives p=12 liters. Ans: 12 liters of the 20 % acidic solution was needed
10:46. A student adds 60 liter of water to an alcohol solution of 20 liters and 28% alcohol . So the student adds 0 liter of alcohol to a solution that has 5.6 liters of alcohol (20 L x 28%) TO PRODUCE 80 liters (60 +20) of an unknown PERCENTAGE of alcohol. Let that unknown % of alcohol in the mixture = R so R% (80) liters of alcohol was in the final solution. Therefore 80 R% = 5.6 + 0 or 80 R= 5.6 or R% = 5.6/80 = 56/800 or 7/100 or 7%. Answer: the solution now contain just 7% of alcohol.13:49
