Blonde_At_Heart

I actually received two kinds of bubble wrap: one that was colored yellow (my favorite color) and then this toy thing here. So, unless the battery runs out... hehe. But thanks for the link - I'll have to send that to the friends that gave me the bubble wrap.

Okay, just to make sure I've got it right, the next problem is this: (using the same equation 2 NO + O2 -> 2 NO2) How many molecules of O2 are needed to make 3.76 x 10^22 molecules NO2? Would it be like this: [math]\frac{3.76 \times 10^{22} \mbox{molecules NO2}}{1} \times \frac{1 \mbox{mol}}{6.022 \times 10^{23} \mbox{molecules NO2}} \times \frac{34.014 \mbox{g NO2}}{1 \mbox{mol}} \times \frac{1\times 31.998 \mbox{g O2}}{2 \times 34.014 \mbox{g NO2}} \times [/math] (same problem, just too long to fit in one [math ]) [math]\frac{1 \mbox{mol}}{31.998 \mbox{g}} \times \frac{6.022 \times 10 ^ {23} \mbox{molecules O2}}{1 \mbox{mol}}[/math] And I know that is overly complicated and almost everything cancels out, but I was only taught one way to do this. I will simplify how I do it later - all I want to know is if that is one way of getting the correct answer. So the molecules NO2 would cancel, the mol would cancel, the 34.014g NO2 would cancel, the 31.998 g would cancel, the mol would cancel, and the 6.022 x 10^23 would cancel. Leaving: [math]\frac{3.76 \times 10^{22} \mbox{molecules}}{2}[/math] Which is [math]1.88 \times 10^{22} \mbox{molecules O2}[/math], right? EDIT: PS, I hope you don't mind that I used your post above to learn how to use the LaTeX, Cap'n Refsmmat. EDIT, again: Oh, I see how I could have simplified the entire equation. I only had ten minutes to learn how to do this chapter and another, so I only learned how to apply the coefficients by grams. Duh, looking back, I could have saved myself a lot of work if I had realized it worked with molecules as well. Oh well, it took so long to write that, I think I'll leave it.

Oh, I see. Thank you so much! That helps a ton!

So, would it be something like this? 27.7 g O2 x 2 (34.014) g NO x 1 mol x 6.022 x 10^23 __________1 (31.998) g O2__34.014g___1 mol (Ignore the underscores in denominator) So the g O2 would cancel, then the g for NO, and finally the mol? Or is that not the correct way to set it up? 31.998 is the molar mass of O2, and 34.014 is the molar mass of NO

I have to do an assignment that deals with stoichiometry. I know it's basically just converting from one thing to another, but a problem baffles me. I have to change from grams to molecules, and I can't remember what kind of conversion that is. The problem says this: Using the equation provided: 2 NO + O2 -> 2 NO2 How many molecules of NO will react with 27.7 g O2? I know how to set up the problem, but I don't know what conversion factor to use to change from grams to molecules, or even to moles if that's possible. Any tips would be greatly appreciated

Oh, that makes perfect sense. Thank you so much!

Hello, I am Blonde_At_Heart, and I like science. I've only taken Biology and Chemistry, but they fascinate me for some reason. My favorite of the two has to be Biology (I prefer learning concepts over math). I stumbled upon this site looking for help doing one of my chemistry papers. Something interesting about me: I received bubble wrap for Christmas, and thought that was amazing.

I have to decide whether the equation is: Synthesis Single Displacement Double Displacement Decomposition Combustion I know what to do for most of them, but one question baffles me. I already filled in the products and balanced the equation (hopefully correctly), and just need to know what type of reaction this is. 2 HCl + CaCO3 -> CO2 + H2O + CaCl2 I know it isn't synthesis or combustion, and single displacement doesn't sound right. I think it may be double displacement, but decomposition (though there are two reactants) sounds like the most probable answer. Any help would be greatly appreciated