Can someone help me with this chemistry problem please??
822185
PV=nRT\nP=1.25 atm\nV=5l\nT=27C=300K\nR=0.0820574587 latm/K/mole\nn=PV/RT=(1.25atm*5l)/(300K*0.0820574587 l atm/K/mole)\nn=.254moles\n\n.254moles weighs 10.13g\n10.13g/.254moles=39.9g/mole
73319
the earlier equation is not balanced.\nCaO + 2HCl --------> CaCl2 + H2O
539827
(a) 2 NaClO3 -> 2 NaCl + 3 O2\n\n(b) 3 MnCl2 + 2 Al -> 3 Mn + 2 AlCl3\n\n(c) C4H10 + 13/2 O2 -> 5 H20 + 4 CO2 \n                  or\n(c) 2 C4H10 + 13 O2 -> 10 H20 + 8 CO2\n\nBalancing is quite simple for these simple equations. You just have to see the left hand side pick up an element, see how many atoms of that element are there in it and multiply the R.H.S compound containing that element with a number which will be equal to no. of atoms in the l.h.s. Then again repeat the procedure by picking another element.....see it's quite difficult to explain you with writing aneways i hope it helps u :)
195778
1.  C(s)+O2--> CO2(g) - 393.5 kj\n2.  H2(g)+1/2 O2(g)--> H20 -285.8 kj\n3.  2C2H2(g)+5O(g)-->4CO2(g)+2H2O(... -2598.8\nCALCULATE the value for the reaction 2C+H2-->C2H2\n\nFirst balance eqn 3 with 5O2 (g) then reverse it so that you have 2C2H2(g) on the RHS. Of course, you will change the sign of the eqn as well. Now halve it so you have one C2H2 instead of 2. Let's call the new eqn number 4\n\nSecond, double the 1st eqn. Let's call it eqn 5\n\nLastly, add eqns 3, 4 and 5:\n\n4 2CO2(g)+H2O --> C2H2(g)+5O2(g) +1299.4 kJmol-1\n5 2C(s)+ 2O2--> 2CO2(g) -787.0 kJmol-1\n3 H2(g)+1/2 O2(g)--> H20 -285.8 kj\n----------------------------------------------\n  2C(s)+ H2(g) --> C2H2(g) + 226.6 kJmol-1
677826
1) is already balanced... to be more precise\n2C2H6 + 7O2 --> 4 C02 + 6H2O\n\n2) Fe2O3 + (3)H2SO4 --> Fe2(SO4)3 + (3)H2O
425080
NaCO3(s) -->  2Na+(aq)  + CO3 2-(aq)
436191
Uh, the rest of the question isn't appearing. But it's an equilibrium solubility problem. Write a reaction:\n\nMgF2 --> Mg+2 + 2F-\n\nKsp = 5.52e-9\n\nFill in the values on the ICE table (I don't know them, the numbers are missing), and set an equation up.\n\nKsp = [F-]^2 * [Mg+2]\n\nSolve with the numbers from the ICE table. Remember that those concentrations in the above equation are the equilibrated concentrations, not the initial.\n\n\nHope this helped.
799504
FeS + 2HBr --> FeBr2 + H2S\nFe2+ + S2- + 2H+ + 2Br- --> Fe2+ + 2Br- + H2S\n2H+ + S2- --> H2S\n\nK2CO3 + CuCl2 --> 2KCl + CuCO3\n2K+ + CO32- + Cu2+ + 2Cl- --> 2K+ + 2Cl- + CuCO3\nCu2+ + CO32- --> CuCO3\n\npeace\nvixklen
733447
Br+H20--->HBr+O\n\nNi(II)O+H20--->Ni+H202\n\nKO2+H2O--->KOH + H2\n\nHope that helps.
813421
Na2CO3(aq) + MgCl2(aq) --> 2NaCl(aq) + MgCO3(aq)
160963
K2SO4 ---> 2K+  +  SO4-\n\nHCl ---> H+  +  Cl-\n\nNaHCO3 ---> Na+  +  HCO3-\n\nNH4OH ---> NH4+  +  OH-\n\nBaCl2 ---> Ba-2  +  2CL-\n\n(NH4)3PO4 ---> 3NH4+  +  PO4-3
890894
C12H22O11+12O2-------12CO2+11H2O+E
775000
2 MnO4-(aq ) + 5 H2C2O4(aq ) + 6 H3O+(aq ) --> 2Mn2+(aq ) + 10 CO2(aq ) + 14 H2O
382458
H2 + Cl2 -> 2HCl\n2Ag + H2S -> Ag2S + H2 \nCH4 + 2O2 -> CO2 + 2H2O\n2HgO -> 2Hg + O2\n3Fe + 4H2O -> Fe3O4 + 4H2
739042
2Al + 6HNO3 ----> 2Al(NO3)3(s) + 3H2(g)
1040448
ask your chem prof or tutor ..it's hard to explain over the internet!
881429
Fe2O3(s) + 3 CO(g) ----> 2 Fe(l) + 3 CO2(g)\n\nCount up the oxygens; the left side and right side oxygens should match.