Question on Thermochemistry - Help!?
When potassium nitrate dissolves in water the temperature of the solution gets colder. Which statement about this reaction is correct? a. The reaction is exothermic and q is positive. b. The reaction is exothermic and q is negative. c. The reaction is endothermic and q is negative. d.The reaction is endothermic and q is positive. Any help would be amazing!
I have TWO thermochemistry Questions! HELP!?
Problem 1 Heat = mass x specific heat x T change 303,000 J = 49,000 g x 4.18 J/gC x T change T change = 303,000 J / (49,000 g x 4.18 J/gC) = 1.48 C T change = T(final) - T(initial) T(final) = T(initial) + T change = 27.0 + 1.48 = 28.5 C Problem 2 Heat = mass x specific heat x T change Heat = 48,000 g x 2.42 J/gC x 4.0 C = 465,000 J = 465 kJ
Thermochemistry question help me please?
You're going to need to make 4 calculations 1: ice at -38 to ice at 0. heat = ΔT x mass x SHC = 38° x 83g x 2.09J/g° = 2: ice at 0 to water at 0 (melt the ice) heat = mass x heat of fusion = 83g x 334J/g = 3: water at 0 to water at 69 heat = ΔT x mass x SHC = 69° x 83g x 4.18J/g° = 4: add up the three heats you just calculated. multiply by [1kJ/1000J] to convert to kJ.
ThermoChemistry Question, Easy Points?
How many grams of methane [CH4(g)] must be combusted to heat 1.00 kg of water from 25 C to 90 C, assuming that H2O as a product and 100% efficiency in heat transfer? This is how far I got: So, calorimetry can be used to solve this. q is the enthalpy q= (mass) (specific heat) ( change in temperature) so, the change in temperature is 65 degrees C The specific heat of water is 4.184 The balanced equation is CH4 + 2 O2 = 2 H2O + CO2 Remember, we are finding the number of grams that are needed. so I also used the fact that bond enthalpies come into play here. Sum of bond enthalpies of products - Sum of bond enthalpies of reactants = The total enthalpy of the reaction. Bond enthalpy of H2O is -285.8 kJ/mol " " " " " " " " " " " CO2 is -393.5 kJ/mol " " " " " " " " " " " CH4 is 74.8 kJ/mol Now I need work and calculations. I gave you a lot of information that I calculated so far. Proceed. Edit: Okay, I've discovered the answer, but I just need to know the work that shows how to get there. (The answer is in the back of the book) The correct mass needed is 4.90 g of CH4. Plz show all work on how to get that value.
Chemistry Question- Thermochemistry?
Heat gained by cooler solution = heat lost by hotter solution (mass cooler solution)(sp.ht. H2O)(Tf - Ti) = (mass hotter solution)(sp.ht. H2O)(Ti - Tf) (30.0 g)(1.00 cal/g K)(Tf - 280 K) = (50.0 g)(1.00 cal/g K)(330 K - Tf) 30.0 (Tf - 280) = 50.0 (330 - Tf) 30.0 Tf - 8400 = 16500 - 50.0 Tf 80.0 Tf = 24900 Tf = 311 K
Can someone help with this thermochemistry question?
When the equilibrium constant K at 25oC for 2 NOCl(g) = 2 NO(g) + Cl2(g) is calculated, using thermodynamic data from your textbook (use delta G values) and the fact that DGof = + 66.05 kJ/mol for NOCl(g), the value obtained is approximately a. ) 6.6 x 10-8 b. ) 16 c. )1.1 x 10-86 d. ) 9.2 x 1085 e. ) 2.4 x 107
Chemistry question - thermochemistry...?
a)the reaction's heat evoled is per mole... i) we have 1/3 of the amount of molecules present, thus 1/3 the heat evolved. ii) the reaction is the same only in the oposite direction, when this happens we just put an negative sign in front of the ^H due to the fact it must take the same amount fo heat. b) solve using the ideal gas law PV=nRT V= 150 cm^3 P= 758 mmHg R= gas constant T= 45 C just convert these and solve for the number of moles you have n. then simply -92.2 KJ/mol * 1mol/ 2 mol NH3 that will be your answer for b
Thermochemistry, calorimetry questions?
In a constant volume bomb calorimeter, the combustion of 0.4921 g of an organic compound with a molecular mass of 46.07 amu causes the temperature in the calorimeter to rise from 25.000 oC to 30.302 oC. The total heat capacity of the calorimeter and all its contents is 3576 J oC-1. What is the energy of combustion of the organic compound, DU / kJ mol-1? Pay attention to sign and significant figures! If you wish to use scientific notation, use the "e" format: e.g. 7.31e4 = 73100 or 1.90e-2 = 0.0190.