Reaction BNAHCO3(s)Na20(s)_CO2(g) +H2O(g)Run 1: Initial grams of NaHCO3Theoretical yield (grams) of NażO:Percent yield (Run 1) =Run 2: Initial grams of NaHCO3 =Theoretical yield (grams) of Na2O:Percent yield (Run 2) = DATARecord all masses to the maximum number ofRun 1Run 2sig figs200°C400°C1.Mass of beaker81.06127.9862.Mass of beaker + NaHCO385.45732.2673.Initial mass of NaHCO34.4764.292Mass of beaker plus products afterheating.4.84.3 8030.7425.Mass of product (actual yield)There are three theoretically possible chemical reactions that could occur during the thermal decomposition of bakingsoda.1)sodium bicarbonate (s) → sodium hydroxide (s) + carbon dioxide (g)2)sodium bicarbonate (s) → sodium oxide (s) + carbon dioxide (g) + water (g)3)sodium bicarbonate (s)→ sodium carbonate (s) + carbon dioxide (g) + water (g)

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Answered: Run 2: Initial grams o

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Chemistry

Run 2: Initial grams o

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter5: Thermochemistry

Section: Chapter Questions

Problem 5.101QE: In the 1880s, Frederick Trouton noted that the enthalpy of vaporization of 1 mol pure liquid is...

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Using data for DHof (KJ/mol) provided below, calculate DHoRxn for: N2O(g) + 3H2(g) ---> N2H4 (l) + H2O (l) DHof (KJ/mol): N2O(g) = 33.85 ; H2 (g)= 0 ; N2H4 (l)= 50.4 ; H2O(l) = -285.8
Use the molar DHf° under the formulas to calculate DH°rxn for the equations as balanced: 1. 2B2H6(g) + 3CO2(g) --> 2B2O3(s) + 3CH4(g) ΔH°rxn = _______ kJ ΔH°f = +36 –394 –1274 –75 kJ/mol exo ? endo_thermic 2. 2P2O5 + 2CaC2 -->P4 + 2CaCO3 + 2CO2 ΔH°rxn = _______ kJ ΔH°f = –1505 –59 ___ –1207 –394 kJ/mol exo ? endo_thermic 3. 2Na2CrO4(s) + 10HCl(g) --> 4NaCl(s) + 3Cl2(g) + Cr2O3(s) + 5H2O(l) ΔH°f = –1342 –92 –411 ___ –1140 –286 kJ/mol ΔH°rxn = _______ kJ exo ? endo_thermic
mass of the calorimeter 162.2 g capacity of the calorimeter 1545.6 Mass of the calorimeter with 140 ml of water 271.3 g Water temperature 22 °C Temperature of the mixture 3 °C Mass of the calorimeter + mixture 310.3 g Calculate 11- ∆T H2O = (Tm - T cold H2O) = (Tm - T cold H2O) = 12- ∆T of ice melt = (T mixture - 0 °C) = (T mixture - 0 °C) = 13- ∆T ° C = ∆T cold H2O:14- Mass of water = ( mass of water + 140 ml of water - mass of empty calorimeter) =.15- Mass of ice = ( mass of calorimeter + mixture ) - (mass of water + 140 ml of water) =
The balanced combustion reaction for C6H6C6H6 is 2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ2C6H6(l)+15O2(g)⟶12CO2(g)+6H2O(l)+6542 kJ If 6.100 g C6H66.100 g C6H6 is burned and the heat produced from the burning is added to 5691 g5691 g of water at 21 ∘21 ∘C, what is the final temperature of the water?
Liquid n-Hexane + oxygen react to to form carbon dioxide + water vapor. delta Hr= __? Kj Liquid sodium sulfate reacts with carbon monoxide(g) to form liquid sodium sulfide and carbon dioxide (g). delta Hr =__? KJ
1. Balance the equation P2O5 + H2O -> H3PO4 What is the coeffiecient of P2O5, H2O, H3PO4 respectively? 1A. Balance the equation B10H18 + O2 -> B2O3 + H2O What is the coeffiecient of B10H18, O2, B2O3, H2O respectively? 1B. Nitrogen gas and Hydrogen gas were combined to form Ammonia. Compute for the grams of Nitrogen when Hydrogen is 7.08 mol Atomic weight: Nitrogen 14.0g/mol Hydrogen 1.00g/mol 1C. Nitrogen gas and Hydrogen gas were combined to form Ammonia. Compute for the grams of Ammonia when Nitrogen is 3.09 grams. Atomic weight: Nitrogen 14.0g/mol Hydrogen 1.00g/mol
The reaction SO2(g)+2H2S(g)⇌3S(s)+2H2O(g)SO2(g)+2H2S(g)⇌3S(s)+2H2O(g) is the basis of a suggested method for removal of SO2SO2, a pollutant that irritates airways causing coughing, from power-plant stack gases. The values below may be helpful when answering questions about the process. Substance ΔG∘fΔGf∘(kJ/mol)(kJ/mol) ΔH∘fΔHf∘(kJ/mol)(kJ/mol) H2O(g)H2O(g) −−228.6 −−241.8 H2O(l)H2O(l) −−237.1 −−285.8 SO2(g)SO2(g) −−300.4 −−296.9 SO3(g)SO3(g) −−370.4 −−395.2 H2S(g)H2S(g) −−33.01 −−20.17 S(s)S(s) 0 0 Assume that the partial pressure of sulfur dioxide, PSO2PSO2, is equal to the partial pressure of dihydrogen sulfide, PH2SPH2S, and therefore PSO2=PH2SPSO2=PH2S. If the vapor pressure of water is 24 torrtorr , calculate the equilibrium partial pressure of SO2SO2 (PSO2PSO2) in the system at 298 KK. Express the pressure in atmospheres to two significant figures. Kp= 8*10^15
[tot] Mn(s) + O2(g) → MnO2(s) ΔH(tot) = ___________? [1] 2MnO2(s) → 2MnO(s) + O2(g) ΔH1 = +264 kJ [2] MnO2(s) + Mn(s) → 2MnO(s) ΔH2 = –240 kJ [tot] = __×[1] __×[2]: ________________→________________ ΔH(tot) = __×ΔH1 __×ΔH2 = ______________kJ
Given that H2(g)+F2(g)⟶2HF(g) Δ?∘rxn=−546.6kJ 2H2(g)+O2(g)⟶2H2O(l) Δ?∘rxn=−571.6 kJ calculate the value of ΔHrxn∘ for 2F2(g)+2H2O(l)⟶4HF(g)+O2(g) Δ?∘rxn= ?kJ
2 LiOH(s) → Li2O(s) + H2O(l) ΔH° = 379.1 kJLiH(s) + H2O(l) → LiOH(s) + H2(g) ΔH° = -111.0 kJ2 H2(g) + O2(g) → 2 H2O(l) ΔH° = -285.9 kJ Compute ΔH° in kJ for 2 LiH(s) + O2(g) → Li2O(s) + H2O(l)
H2SO3 = 0.2556 M, Ka1 = 1.6 x 10-2 , Ka2 = 6.4 x 10-8 NaOH = 0.3106 M please show all work all information needed is here
A. Calculate the moles of anhydrous (dry) KAl(SO4)2 that were present in the sample. Step 1: Find the mass of the dry anhydrous salt show work: Final weighing - mass of pan = mass of anhydrous salt Step 2: Find the molar mass of KAl(SO4)2 and show work here Step 3: Find the moles with work shown: grams of the dry salt divided by the molar mass = moles B. Calculate the ratio of moles of H2O to moles of anhydrous KAl(SO4)2. Note: Report the ratio to the closest whole number. Show work here: Take the value of the moles of water and divide by the moles of the salt. Round to the nearest whole number C. Write the empirical formula for the hydrated alum, based on your experimental results and answer to Question B. Hint: if the ratio of moles of H2O to moles of anhydrous KAl(SO4)2 was 4, then the empirical formula would be: KAl(S04)2•4H20. Show your work. the image uploaded is the table that is needed to answer these questions