Hi, I am not sure if I am understanding this incorrectly but in my calculator when I input (0.035)(0.035)/(0.015) I am getting 0.81666...7, I might have missed a step or misunderstood, could you explain if I missed a calculation or misunderstood the calculation pls. Btw, love your videos, never have I understood chem so well, thank you using your time to make these videos.
I want to thank you for making this equilibrium video as well as the concepts of how to get to equilibrium by demonstrating the “seesaw” visual. I’m the type of student (sometimes, not always) who needs that extra explanation or “visual” to get the point of steps. From the bottom of my heart, I can’t thank you and the teacher who put the extra effort for students such as my self to get the grasp of what we’re actually learning. It’s teachers like you and alike who are the real MVP’s! ❤ God bless you!
H MGloria, I am so glad the videos were helpful. It makes me very happy. I am proud of your extra work and effort searching for videos to better understand the content. God bless you!! Thanks:)
K is not unitless is it? Here wouldn’t it have units of M? In general, it would depend on the degree of the numerator minus the degree of the denominator.
It is a ratio of product and reactant concentrations which are both M. You are correct, based on exponents, mathematically it can have a final M to the x power. However, because it is concentration divided by concentration it is considered unitless. It is just a ratio of concentrations at equilibrium for K. Thanks!
@@chiararomano1818 Hi Chiara, It is one of the main differences between the equilibrium constant (K) and rate constant (k). It is so great you noticed that!
You’re so helpful! I have an exam in a couple of days and you are a lifesaver! Thanks for uploading.
Hi Katie, I’m so glad! Good luck on your test😊
Hi, I am not sure if I am understanding this incorrectly but in my calculator when I input (0.035)(0.035)/(0.015) I am getting 0.81666...7, I might have missed a step or misunderstood, could you explain if I missed a calculation or misunderstood the calculation pls.
Btw, love your videos, never have I understood chem so well, thank you using your time to make these videos.
Hi Charank, You are correct! Thank you for catching that. I must have made a calculator error. Thanks again!
@@OldSchoolChemistry No problem, thank you so much for clarifying!
@@OldSchoolChemistryso it’s going to be reactant favoured? Because it’s less than 1?
@@llennzo Yes:) Well done.
I want to thank you for making this equilibrium video as well as the concepts of how to get to equilibrium by demonstrating the “seesaw” visual. I’m the type of student (sometimes, not always) who needs that extra explanation or “visual” to get the point of steps. From the bottom of my heart, I can’t thank you and the teacher who put the extra effort for students such as my self to get the grasp of what we’re actually learning. It’s teachers like you and alike who are the real MVP’s! ❤ God bless you!
H MGloria, I am so glad the videos were helpful. It makes me very happy. I am proud of your extra work and effort searching for videos to better understand the content. God bless you!! Thanks:)
Thank you so much! Your videos are great and so helpful
Hi Fana, I am so glad! Thank you for your comment;)
I think I might have done the calculations wrong; I keep getting .082 for my equilibrium constant value
Hi Ian, you are correct! Thank you for catching that. I somehow accidentally reciprocated the answer. Yes it is 0.082. Thanks!
@@OldSchoolChemistry Thank you so much btw your awesome, without you I don't think I would be succeeding in chem 2!
Good luck in your class 😊
K is not unitless is it? Here wouldn’t it have units of M? In general, it would depend on the degree of the numerator minus the degree of the denominator.
It is a ratio of product and reactant concentrations which are both M. You are correct, based on exponents, mathematically it can have a final M to the x power. However, because it is concentration divided by concentration it is considered unitless. It is just a ratio of concentrations at equilibrium for K. Thanks!
@@OldSchoolChemistry Thanks. I didn’t know that. The constants for reaction rates carry units, I just assumed the equilibrium constants would as well.
@@chiararomano1818 Hi Chiara, It is one of the main differences between the equilibrium constant (K) and rate constant (k). It is so great you noticed that!
Awesome videos
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thank you so much 😘🥰 you are a beautiful soul
Hi Jennipher, thank you for your comment:) I appreciate you!💗
@@OldSchoolChemistry they brought the exact question in the exam yesterday 😍😍💃💃💃💌
@@jennipherchanda6266 Hooray! And you were completely prepared. Way to go!!!!!