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Chemistry— Stoichiometry?
When 0.2235g of strontium reacted with 25.00mL of 1.168 M HCl, 63.18g of H20 (water) was displaced by the H2 gas produced. The temperature in the reaction flask was 27.0 degrees C and the barometric pressure was 777 Torr. The molar mass of strontium is 87.62g/mol. At 27.0 degrees C the vapor pressure of water is 27.0 Torr and its density is 0.9965g/mL. When the final reaction mixture was titrated with 0.6137M NaOH, the endpoint was reached when 39.25mL of base were added.
d. Calculate the number of moles or unreacted HCl left in the reaction mixture.
In other parts, I found the moles of gas collected to be 0.00252 mol H2, the volume of gas collected to be 63.40 mL, and moles of HCL used to be 0.005102 mol HCl. How would I set up part d?
2 AnswersChemistry6 years agoChemistry: Stoichiometry?
When 0.2235g of strontium reacted with 25.00mL of 1.168 M HCl, 63.18g of H20 (water) was displaced by the H2 gas produced. The temperature in the reaction flask was 27.0 degrees C and the barometric pressure was 777 Torr. The molar mass of strontium is 87.62g/mol. At 27.0 degrees C the vapor pressure of water is 27.0 Torr and its density is 0.9965g/mL. When the final reaction mixture was titrated with 0.6137M NaOH, the endpoint was reached when 39.25mL of base were added.
c. Calculate the number of moles of HCl used as a reactant.
How would I do this? I have the balanced equation set up ( Sr + 2HCl ----> SrCl2 + H2), but I'm stuck after that.
Any help would be greatly appreciated, Thanks!
1 AnswerChemistry6 years agoGrams required and grams produced--- Help would be appreciated!?
Carbon dioxide is considered to be the green house gas of greatest concern with respect to global warming. A considerable amount of carbon dioxide is produced upon the combustion of gasoline. If you assume that gasoline is mostly octane, C8H18, then the combustion reaction is
2 C8H18(l) + 25 O2(s) ---> 16 CO2(s) + 18 H2O(l)
a) If you burned 1352 g (which is about 1 quart) of octane, C8H18, how many grams of oxygen, O2, are required?
b) How many grams of CO2 is produced?
1 AnswerChemistry6 years agoBio Multiple Choice Question?
Why would a frog risk becoming vulnerable to predators by croaking to find a mate?
(A) The frog knows it needs to reproduce sexually to have variable offspring
(B) Only frogs that croak are able to leave offspring in the next generation
(C) Males aren’t able to reproduce asexually like females so they must find a mate
I know it's not (A) because this is my second time doing this review-quiz-thing and it told me that it was wrong. I'm just trying to understand whether (B) or (C) are the correct answers. I could see it being the former because the croak is like a mating call and announces to females that they want to do the do. But, after a quick search online, I got results saying some frogs, like the Tungara frog, are asexual; this supports choice (C). We were actually shown a video about the Tungara frog so I feel like it could be (C). The females can only produce via parthenogenesis, though. Help, please? Haha. :)
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But, we've started hanging out alone a lot more. Recently, we went to a sporting event and a movie by ourselves. Usually, he makes sure to have more people, but this time, he almost tried to make sure it was just us. We talked about past relationships and what each of us would want from the future.
Twice in the past it's come out (from the loose lips of a former friend) that I liked my guy best friend. Honestly, I wasn't ready for a relationship at those points and that's why I didn't say anything. He has said that he "knew" those times. I thought it was weird because at the height of me liking him was when we were the closest. Also, for my birthday, he painted a picture of two cats forming a heart with the spots on their backs. I jokingly asked for it and he actually gave it to me.
Sorry for the long and rambling post, but if anyone has any advice or observations, that would be neat. I can provide other info too, if you want. :)
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So, do you think I should pursue anything out of this, or not?
Please and Thank You :)
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