Chemistry/Physics Question.?
820189
Because the volume, pressure, and temperature are all related by Boyles Law and the Ideal Gas Law.\n\npv=nRt\n\nPressure times Volume equals the Number of molecules times the Gas Constant time the temperature.
870625
each of the particles would have 1/2 the energy of the spring, 30 j each.  They would move at different speeds, c moves slower than d, but energy 1/2 mv^2 would be the same.
2247
And this is chemistry how??
1100440
This site should help you.\n\nhttp://sparknotes.com/chemistry\n\nGood luck.
933519
Part A:  equilibruim to the right because more co3 added\n\nPart B:  Equilibrum to left becausemore co2 added\n\nPart C:  Equilibrium to right to replace removed h20\n\nEquilibruim dynamic because although appears stable ions constantly moving back and forth just at the same time
1142668
Problem \ntake D as Delta(change)\nEstimate the change in enthalpy, DH, for the following reaction:\n\nH2(g) + Cl2(g) --> 2 HCl(g) \n\nSolution \n\nTo work this problem, think of the reaction in terms of simple steps:\n\n\nFirst, the reactant molecules, H2 and Cl2, break down into their atoms\nH2(g) --> 2 H(g)\nCl2(g) --> 2 Cl(g)\n\n\nNext, these atoms combine to form HCl molecules\n2 H(g) + 2 Cl(g) --> 2 HCl(g) \n\nIn the first step, the H-H and Cl-Cl bonds are broken. In both cases, one mole of bonds is broken. When we look up the single bond energies for the H-H and Cl-Cl bonds, we find them to be +436 kJ/mol and + 243 kJ/mol, therefore for the first step of the reaction:\n\nDH1 = +(436 kJ + 243 kJ) = +679 kJ\n\nBond breaking requires energy, so we expect the value for DH to be positive for this step.\n\nIn the second step of the reaction, two moles are H-Cl bonds are formed. Bond breaking liberates energy, so we expect the DH for this portion of the reaction to have a negative value. Using the table, the single bond energy for one mole of H-Cl bonds is found to be 431 kJ:\n\nDH2 = -2(431 kJ) = -862 kJ\n\nBy applying Hess's Law, DH = DH1 + DH2\n\nDH = +679 kJ - 862 kJ\n\nDH = -183 kJ\n\n\nAnswer\n\nDH = -183 kJ
801048
I think this belongs in the Chemistry section.  I'm not a Chemistry buff, but if someone here is, I'm sure you'll get your answer.
949133
OMG!\ni have the same question!\nmine is:\nA piece of an unknown metal has a mass of 20.0 g and at 100.0 C is placed in a styrofoam cup calorimeter containing 50.7 g of water at 22 C. If the final temp is 27.2 C, calculate the specific heat capacity of the metal. Assume that all the energy lost by the metal is gained by the water.\n\nok, haha fine, not the SAME question, but close enough. But, i'm uber confused. ><
844758
Using the Michaelis-Menten equation.\n\nMore details on teh attached link\n\nhttp://www.le.ac.uk/by/teach/biochemweb/tutorials/michment2print.html\n\nGood luck
735831
Did you put your entire homework assignment on here? I noticed you posted a bunch of science questions just now.\nIf you don't have your textbook handy [you really should be looking up these answers so that you can remember and understand them when you have your test/exam], go to the links below and look them up. The 1st is a down loadable physics textbook. The 2nd has animations of physics.
991058
mass =concentration\n-------\nvolume
294164
Thw formular to use here is Q = mcT where Q = quantity of heat, m = mass, c = specific heat capacity, and T is the change in the temperature, thus here Q = 2.0g*(30-25)*0.90 = 9J. Notice that I sed Joules as the unit.\n\n2.For this one, a law which i cant remember the name right now is nedeed. Here we have C(s)+O2(g)=CO2 Enthalphy -390KJ\nand also Mn(s)+O2(g)=MnO2(s) Enthalphy = -520KJ\nFor this law to be appplied, we have to write the second equation the other way to become MnO2= Mn+O2 Enthalphy = 520KJ. Notice that i changed the sign when i reversed the equation. Then O2 could be canceled out on the left side of the first equation and on the right side of the second equation. This was we are adding. Finally adding, we get MnO2(s)+C(s) = Mn + CO2 Enthalphy = -390KJ+520KJ = 130KJ.\nand thats the correct answer
944088
CS2 (g) +4H2 (g) <----> CH4 (g) +2H2S(g) \n\na) pressure of CS2 increase,    -  NO\nb) pressure of CH4 decrease,   - NO\nc) pressure of H2 decrease,     -  YES\nd) pressure of H2S decrease,   -  NO\ne) all the pressures will decrease  -  NO
64647
2x (G2 + M2 ---> 2GM)      DH=-280.0*2= -560.0 kJ\n2x (L + 2M2 ---> LM4)      DH=-350.0*2= -700.0 kJ\n   (L2G4 ---> 2L + 2 G2)   DH=122.0*(-1)=-122.0 kJ (change sign when flip over)\n\nAdd them together, we'll have:\n2G2 + 2L + 6M2 + L2G4 ---> 4GM + 2LM4 + 2L + 2G2\nthen cancel 2G2 and 2L on both side, we end up with:\nL2G4 + 6M2 ---> 2LM4 + 4GM\nthe enthalpy of this reaction is, therefore the sum of each enthalpies calculated above.\nDH=(-560.0) + (-700.0) + (-122.0)= -1382.0 kJ (exothermic process)\nhope that'll help!
62030
Protons and neutrons.
898375
I know the answer is in my basic physics book.  I would have to look it up.  You should too.
291938
1. electron\n2. ionic\n3. molecules of compounds\n4. decomposition (not sure)