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fandomhigh2006-02-13 09:29 am
equalsmcsquared.livejournal.com) wrote in fandomhigh2006-02-13 09:29 am
Biology (2nd Period) / Chemistry (4th Period)
"Good morning, everyone. I hope you had a good weekend. If not, oh well, I don't think I'll be making things any better for you." She smiles pleasantly, despite her words.
"Feel free to help yourselves to the candy and cupcakes."
"Now, today we're going to begin discussing animal taxonomy, or how we classify various types of animals in the animal kingdom."
She passes around an introductory worksheet as well as giving an in-depth lecture on the principles of classification.
"Homework: choose any animal and give me the basic classification. Bonus points for genus and species."
"Today, we're going to discuss limiting reagents. I know this stuff seems tedious but it is vital when doing lab work. You don't want unbalanced solutions. Trust me when I say that there is no need to live through an explosion."
Sometimes when reactions occur between two or more substances, one reactant runs out before the other. That is called the "limiting reagent". Often, it is necessary to identify the limiting reagent in a problem.
Example: A chemist only has 6.0 grams of C2H2 and an unlimitted supply of oxygen and desires to produce as much CO2 as possible. If she uses the equation below, how much oxygen should she add to the reaction?
2C2H2(g) + 5O2(g) ---> 4CO2(g) + 2 H2O(l)
To solve this problem, it is necessary to determine how much oxygen should be added if all of the reactants were used up (this is the way to produce the maximum amount of CO2).
First, we calculate the number of moles of C2H2 in 6.0 grams of C2H2. To be able to calculate the moles we need to look at a periodic table and see that 1 mole of C weighs 12.0 grams and H weighs 1.0 gram. Therefore we know that 1 mole of C2H2 weighs 26 grams (2*12 grams + 2*1 gram).
Then, because there are five (5) molecules of oxygen to every two (2) molecules of C2H2, we need to multiply the result by 5/2 to get the total molecules of oxygen. Then we convert to grams to find the amount of oxygen that needs to be added:
"Homework: Do the odd problems on pgs. 46 - 53."
"Feel free to help yourselves to the candy and cupcakes."
"Now, today we're going to begin discussing animal taxonomy, or how we classify various types of animals in the animal kingdom."
She passes around an introductory worksheet as well as giving an in-depth lecture on the principles of classification.
"Homework: choose any animal and give me the basic classification. Bonus points for genus and species."
"Today, we're going to discuss limiting reagents. I know this stuff seems tedious but it is vital when doing lab work. You don't want unbalanced solutions. Trust me when I say that there is no need to live through an explosion."
Sometimes when reactions occur between two or more substances, one reactant runs out before the other. That is called the "limiting reagent". Often, it is necessary to identify the limiting reagent in a problem.
Example: A chemist only has 6.0 grams of C2H2 and an unlimitted supply of oxygen and desires to produce as much CO2 as possible. If she uses the equation below, how much oxygen should she add to the reaction?
2C2H2(g) + 5O2(g) ---> 4CO2(g) + 2 H2O(l)
To solve this problem, it is necessary to determine how much oxygen should be added if all of the reactants were used up (this is the way to produce the maximum amount of CO2).
First, we calculate the number of moles of C2H2 in 6.0 grams of C2H2. To be able to calculate the moles we need to look at a periodic table and see that 1 mole of C weighs 12.0 grams and H weighs 1.0 gram. Therefore we know that 1 mole of C2H2 weighs 26 grams (2*12 grams + 2*1 gram).
Then, because there are five (5) molecules of oxygen to every two (2) molecules of C2H2, we need to multiply the result by 5/2 to get the total molecules of oxygen. Then we convert to grams to find the amount of oxygen that needs to be added:
"Homework: Do the odd problems on pgs. 46 - 53."