Recent content by Macroer

Thanks! I would have never found that. :smile:

Homework Statement A uniform rod of length 3L is pivoted as shown and is resting at a horizontal position. What is the acceleration of the center of mass of the rod the instant it is released? The diagram is added to the attachment. Homework Equations tnet=IaangularThe Attempt at a Solution...

So... ksp=[Sr2+][SO42-] 3.4x10^-7=[Sr2+][SO42-] 3.4x10^-7=[0.1M][SO42-] [SO42-]=3.4x10^-6M Now in Pb2+ 1.8x10^-8=[Pb2+][SO42-] 1.8x10^-8=[Pb2+][3.4x10^-6M] [Pb2+]=5.29x10^-3M Is this how i do it, or do i have to set up an ICE table?

Homework Statement 1. A 1L solution contains 0.1M of each Pb+2,Ca+2, and Sr+2. Which ion precipitates last as Na2SO4 is slowly added with no change in volume? 2. What is the concentration of the ion that precipitates first when the second ion precipitates? Given: Ksp PbSO4=1.8x10^-8 Ksp...

I don't understand where you got the 2nd chlorine atom from

I* + CH3-CCl=CH-CH3--> I-CH2-CCl*-CH2-CH3 And i am not sure what the "end groups" you are referring to are.

Initiation: I* + CH2=CCl-CH2-CH3--> I-CH2-CHCl*-CH2-CH3How is that for initiation step? where I is initiator and * is radical sign

Homework Statement How would the polymer of 2-chloro-2-butene look when formed by addition polymerization Homework Equations The Attempt at a Solution I have attached my solution. Please tell me if i am right/wrong and why if it is wrong.

I got a followup question concerning this: By calculating the volume after we used the same total volume of solution(50mL and 100mL). However, hydrogen gas was lost, doesn't that lower the volume of the solution?

Nope, the 50mL one, had a greater change in concentration than the 100mL one, which produces greater error because rate law is only accurate for initial concentrations of reactants?

so for 50mL, and 2M solution 2mol/L*(0.05L)=0.1mol amount used =0.04mol (.1-0.04mol)/0.05L=1.2 mol/L and for 100mL and 2M solution 2mol/L*(0.1L)=0.2mol (0.2mol-0.04mol)/0.1L)=1.6mol/L

50mL 1.33mol/0.05L=26.6mol/L 100mL 2.7mol/0.1L=27mol/L err, those numbers seem big, must of done a calculation wrong.

Amount of moles used=0.04 mol For 50 mL 50g/36.46 g/mol=1.37mol-0.04 mol=1.33mol remaining for 100mL 100g/36.46g/mol=2.74mol-0.04 mol=2.7mol remaining

Don't know the density of HCl, but assuming it was the same as water. 1.5g=1.5mL in 50mL amount left=50mL-1.5mL =48.5 in 100mL amount left=100mL-1.5mL =98.5mL

Mg + 2HCl -> MgCl2 + H2 0.5g/24.31g/mol * 2 mol/1 mol * 36.46 g/mol=1.5 g HCl