2:46 if you look down in, you're able to see with the two metal spatulas as it heats up... _proceeds to hold your attention until __3:18__ where you discover he was never gonna say anything_
Awesome ! Thank you! I was confused. We did the experiment and now i'm trying to do the calculations.... was completely lost!!!! You video a huge help!!!! Hope you have more!!!
Just a simple question at 7:58 If we look at the 3 chemical reactions where does the heat go in the chemical equation will it go on the left side making it an Endothermic reaction or right side making it an Exothermic
Took the weight of the baking soda then converted to moles using its molar mass then by using molar ratio he converted to how many moles of NaOH,Na2CO3, and NaOH would have been produced, then he converted moles to grans
A great follow up would be to test both with MgSO4. I am doing this on my own atm since I was some soda ash for my fishtank and it is much cheaper to make it myself.
this is a flawed experiment from the beginning. the baking soda should be desiccated before use to remove any water. and in order to get the true mass, he should have weighed the crucible and test tube before adding the baking soda, and then weighing again after adding. doing it his way, the data are distorted by the losses on the weighing boat and spatula.
Now how in he hell does the dude know what all those chemicals weigh? Did he just memorize those? So what, he weighs the end product to determine what is converted to?
You are able to determine the mass of the end products using stoichiometry. If you wanted to find the mass of sodium carbonate produced, you would multiply the mass of the baking soda you use (2.00g) by one mole over the mass of one mole of Baking soda (84.01g). You then need to determine how many moles of sodium carbonate are produced per mole of Baking soda, which is one mole of Na2CO3(sodium carbonate) per 2 moles of NaHCO3(baking soda). Then you multiply this ratio (1 mol / 2 mol) by the mass of one mole of NaCO3 (103.99g) over one mol. This gives you the equation 2.00g NaHCO3 x (1 mol / 84.01g) x (1 mol / 2 mol) x ( 103.99g / 1 mol ) which simplifies to 1.26 grams of Sodium Carbonate. As to whether he just memorized the values, more than likely yes, as there are very few people capable of doing that kind of math in their heads that quickly.
2:46 if you look down in, you're able to see with the two metal spatulas as it heats up...
_proceeds to hold your attention until __3:18__ where you discover he was never gonna say anything_
lol
Awesome ! Thank you! I was confused. We did the experiment and now i'm trying to do the calculations.... was completely lost!!!! You video a huge help!!!! Hope you have more!!!
I wish my teacher was this good at teaching.
Just a simple question at 7:58 If we look at the 3 chemical reactions where does the heat go in the chemical equation will it go on the left side making it an Endothermic reaction or right side making it an Exothermic
Its an endothermic reaction. Heat needs to be constanty "added" to the sodium bicarbonate for the decomposition to keep going.
where did you get your numbers for the masses of Na2O, Na2CO3 and NaOH
solum1000 molecular weight...
He used the stoichiometry method to find the number of mass they have
Took the weight of the baking soda then converted to moles using its molar mass then by using molar ratio he converted to how many moles of NaOH,Na2CO3, and NaOH would have been produced, then he converted moles to grans
A great follow up would be to test both with MgSO4. I am doing this on my own atm since I was some soda ash for my fishtank and it is much cheaper to make it myself.
Could there be an incomplete or mix of those reactions?
How do you know how long to heat the Sodium Bicarbonate in the crucible?
Noah Robert simple.
You don't
:v
Untill there's no gas coming off
lol thanks man but you're about 11 months too late XD
can the temperature be controlled?
EXCELLENT!!! Simple but effective!!! No hazardous waste👍🏼👍🏼👍🏼👍🏼
Thank you
What is the thing that the test tube is on??
a clay triangle
accurate to 2 decimal points with the weight then transferring it into another container and losses on the spatula ...........
Is the calculation wrong for the first one, because I got .48grams not .953 grams using stoichiometry.
Claudia Lopez your answer is wrong, maybe you missed a step or something
You lied to me FlinnScientific the experiment went bad the bunsen burner exploded in front of me
Excuse me, that means you've made mistake.
"...brought to you by Arm & Hammer"
😂😂😂👍🏼
Guessing can cause death with real chemistry.
what chemical reaction is this
Beautiful reaction 😯
Sodium carbonate breaking down I think?
this is a flawed experiment from the beginning. the baking soda should be desiccated before use to remove any water. and in order to get the true mass, he should have weighed the crucible and test tube before adding the baking soda, and then weighing again after adding. doing it his way, the data are distorted by the losses on the weighing boat and spatula.
Now how in he hell does the dude know what all those chemicals weigh? Did he just memorize those? So what, he weighs the end product to determine what is converted to?
You are able to determine the mass of the end products using stoichiometry. If you wanted to find the mass of sodium carbonate produced, you would multiply the mass of the baking soda you use (2.00g) by one mole over the mass of one mole of Baking soda (84.01g). You then need to determine how many moles of sodium carbonate are produced per mole of Baking soda, which is one mole of Na2CO3(sodium carbonate) per 2 moles of NaHCO3(baking soda). Then you multiply this ratio (1 mol / 2 mol) by the mass of one mole of NaCO3 (103.99g) over one mol. This gives you the equation 2.00g NaHCO3 x (1 mol / 84.01g) x (1 mol / 2 mol) x ( 103.99g / 1 mol ) which simplifies to 1.26 grams of Sodium Carbonate.
As to whether he just memorized the values, more than likely yes, as there are very few people capable of doing that kind of math in their heads that quickly.
he had a paper in his hand
Disliked for wearing eye protection
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