Chemistry help? ksp calculations...?
a) Pb3(PO4)2 <--> 3Pb^(2+) + 2PO4^(3-) As the reaction progresses, due to the stoichiometry of the reaction, +3s of [Pb2^+] and +2s of [PO4^3-] will be made. Ksp = 1.00x10^-54 = [Pb^2+]^3 [PO4^3-]^2 = [3s]^3 [2s]^2 = 108s^5 s = 6.21x10^-12 M b) (6.21x10^-12 mol Pb3(PO4)2 / 1 L) x (811.54 g Pb3(PO4)2/ 1 mol Pb3(PO4)2) = 5.04x10^-9 g Pb3(PO4)2 / 1 L c) Same process as parts a) and b), but with a different reaction. PbCrO4 <--> Pb^(2+) + CrO4^(2-) Plumbous ion concentration is increased by s, chromate ion concentration is increased by s. 2.00x10^-16 = s^2 s = 1.41x10^-8 mol/ L Convert moles to grams to find solubility in terms of grams, then convert liters to mL in order to find the mass of PbCrO4 that will dissolve in 100 mL. (1.41x10^-8 mol/ L) x (323.2 g / mol) x (1 L / 1000 mL) x (100 mL) = 4.56x10^-7 g PbCrO4
Easy Chemistry Gas Laws Question: Calculate the final celsius temperature...?
Calculate the final Celsius temperature required to change 10L of helium at 100K and 0.15atm to 30L at 0.2atm. I got 400K, but I'm not sure if I did it right. Please check my answer and if it's wrong, tell me how you did it. Thanks!
Would someone help me with my chemistry homework on chemical changes?
First you must define your problem! a wick in a candle is usually a cotton or other burnable string general formula C6H10O5. It is an organic compound burning it forms CO2 and H2O thru a series of steps. Such a wick in a candle when ignited burns, melts the wax, and wicks the melted wax to the burning flame where the wax is vaporized and burned, again to CO2 and H2O. Since the burning is at a relatively low temperature it is incomplete and many intermediate compounds [soot, CO] are formed.The wick in a kerosene lantern such as a Colman lantern is made of a ceramic fiber; it does not burn but serves to bring the liquid fuel to the burning flame that must be ignited by a match or other external flame
Stoichiometric Calculations - Reactions - EASY BEST ANSWER!?
F2(g) + 2NaBr(aq) ------------> 2NaF(aq) + Br2(l) M= 100g Mass of Br2(l) produced = 100g / 102.89 g/mol * 1mol Br2 / 2mol NaBr * 159.80g / 1 mol = 77.7 g Br2 produced.
Please Help With Easy Chemistry Question but I am Stupid to Understand lol?
Moles SrF2 dissolved in 100 mL = 1.1 x 10^-2 / 125.62 g/mol=8.8 x 10^-5 [Sr2+]= 8.8 x 10^-5 / 0.100 L=8.8 x 10^-4 M [F-] = 2 x 8.8 x 10^-5/ 0.100 L=1.8 x 10^-3 M Ksp = [Sr2+][F-]^2 = 8.8 x 10^-5 ( 1.8 x 10^-3)^2 =2.7 x 10^-10
Chemistry Question, Help!?
1) You are given the mol of a compound , and are asked to find the mass (usually in grams) of that compound. For these types of conversions, you use the molecular weight of the given compound (g/mol) and set up an equation to figure out the missing number. The molecular weight of Fluorine is around 19g/mol. To set up the equation, its __mol = 19g/mol This way, the mol can cancel out with each other, so we are left with multiplying the remaining numbers to get the grams. 2) Use Avagadro's constant 6.023 x 10^23 molecules/moles. To convert the number of atoms to moles: You need the number of atoms in a substance which is usually given. Ex. 1.65 × 1024 atoms of Magnesium. Then the formula is as follows: # of atoms ÷ (Avogrado's Constant) = # moles 1.65 × 1024 atoms ÷ (6.02 × 1023) = 2.74 moles of ? 3) this is also another question that requires you set up an equation. Given the mass and the average atomic mass, we can set it up 8.2 g X (1 mol/23 grams) = ? ? X 6.02 X 10^23 atoms/mol = answer note: since we are given grams and want mols for the first part of the conversion, we must reverse the g/mol to mol/g so we can cancel out the grams to get the moles, so we could use avogrado's number, which is in atoms/mol to get the number of atoms. You could also do it in one equation, like 8.2 grams X (1 mol/23 grams) X 6.02 X 10^23 atoms/mol = answer 4) Also, similiar concepts to the above, except you have to convert two different compounds and compare them to see which one is heavier with the same amount of mols. Helpful reminder: grams/molar mass= moles moles x avagadros number (6.02 x 10^23 things) = atoms/molecules
PLEASE HELP ME WITH MY CHEMISTRY HOMEWORK!! DUE TOMORROW!?
okay I need help with these problems please :) can you please show me how to do it, and not just the answer. thanks how many molecules are there in .8 grams of krypton? calculate the number of atoms in 60 grams of Ne. calculate the number of oxygen atonms in 2 moles of NaOH. calculate the number of K+ ions produced when 3.0 moles K2SO4 (K little 2 S O little 4) and lastly calculate the number of atoms of 2.5 moles of Ca. THANKS! I really appreciate any answers :)
CHEMISTRY HELP please! by tonight!?
one mole of Fe(OH)2 is the atomic weight of iron+2xatomic wt of O+2xatomic weight of H in grams. Multiply by 2.5 to get your answer.USE THE TABLE OF ELEMENTS. It will be way easier. Anyways these are the answers Each mole of iron (ll) hydroxide weighs 71.85 g/mol. So 2.5 moles would be 71.85 x 2.5= 179.63 grams of FeO Each mole of sodium sulfate weighs 142.04 g/mol .So 2.5 moles would be 355.1 grams of sodium sulfate
Chemistry homework help?
i think you would use plancks law. E = hf h being plancks constant f being frequency also to get from wavelength to frequency you use. and i would think that red is lower energy compared to green yes. f = velocity/wavelength so for red light frequency = speed of light/6.40 x 10-5 cm red f = 4.6875 * 10^14 green f = 5.99 * 10^14 so using plancks law you can clearly see that since energy is proportional to frequceny. Red light is of lower energy than green.
Can I use a calculator while practicing chemistry numericals?
You definitely can use a calculator but you obviously should not use calculator for practice purposes. Calculations might seem boring and time consuming but there is no alternative to that because finally you have to perform these calculations in the test that you are preparing for. And these calculations might seem even more lengthy if you do that at the time of exam directly. Performing calculations while practicing problems would also provide you certain ‘shortcuts and tricks’ that you might use to simplify your calculations at the time of the actual test.Hope this helps!