다운로드 Thermochemistry

다른 한편에서는 원활한 경험을하려면 파일을 장치에 다운로드 한 후 파일을 사용하는 방법을 알아야합니다. APK 파일은 Android 앱의 원시 파일이며 Android 패키지 키트를 의미합니다. 모바일 앱 배포 및 설치를 위해 Android 운영 체제에서 사용하는 패키지 파일 형식입니다.
네 가지 간단한 단계에서 사용 방법을 알려 드리겠습니다. Thermochemistry 귀하의 전화 번호.

1 단계 : 다운로드 Thermochemistry 귀하의 기기에서

아래의 다운로드 미러를 사용하여 지금 당장이 작업을 수행 할 수 있습니다. 그것의 99 % 보장 . 컴퓨터에서 파일을 다운로드하는 경우, 그것을 안드로이드 장치로 옮기십시오.  

2 단계 : 기기에 타사 앱 허용

설치하려면 Thermochemistry 타사 응용 프로그램이 현재 설치 소스로 활성화되어 있는지 확인해야합니다. 메뉴 > 설정 > 보안> 으로 이동하여 알 수없는 소스 를 선택하여 휴대 전화가 Google Play 스토어 이외의 소스에서 앱을 설치하도록 허용하십시오.

3 단계 : 파일 관리자로 이동

이제 위치를 찾으십시오 Thermochemistry 방금 다운로드 한 파일입니다.
일단 당신이 Thermochemistry 파일을 클릭하면 일반 설치 프로세스가 시작됩니다. 메시지가 나타나면 "예" 를 누르십시오. 그러나 화면의 모든 메시지를 읽으십시오.

4 단계 : 즐기십시오

Thermochemistry 이 (가) 귀하의 기기에 설치되었습니다. 즐겨!

개발자 설명

Thermochemistry helps in evaluation of enthalpy or heat release/absorption of a system undergoing numerous temperature changes and phase transitions. The calculations take advantage of known values of heat capacity at constant pressure (Cp) and molar or per gram enthalpy of phase transition. Amount of compound can be defined in grams or moles, in a way that Cp and enthalpy units would match. App provides enthalpy values for each step. Negative enthalpy points to exothermic process – heat release, while positive one to endothermic - heat absorption. °C and K are interchangeable. Calorimetry section provides means for evaluation of the heat capacity of calorimeter and for finding equilibrium temperature of mixed system. Forward arrow button sets the final temperature of the mixture. Backward arrow button sets missing temperature or heat capacity of one of the components. Enthalpy values show heat flow for each component. Example of problems solved by application (screenshots): Problem 1: Calculate the amount of energy required to change 100.0 g of ice at -15.0 °C to steam at 125.0 °C. Known values: Heat of melting = 334.16 J g¯1 Heat of vaporization = 2259 J g¯1 specific heat capacity for solid water (ice) = 2.06 J g¯1 K¯1 specific heat capacity for liquid water = 4.184 J g¯1 K¯1 specific heat capacity for gaseous water (steam) = 2.02 J g¯1K¯1 Solution: 1) Heating of 100.0 g of ice from -15.0°C to 0.0°C: (100.0 g) (15.0 K) (2.06 J g¯1 K¯1) = 3090 J 2) Melting of 100.0 g of ice: (100.0 g) (334.16 J g¯1) = 33416 J 3) Heating of 100.0 g of liquid water from zero to 100.0 Celsius: (100.0 g) (100.0 K) (4.184 J g¯1 K¯1) = 41840 J 4) Evaporations of 100.0 g of liquid: (100.0 g) (2259 J g¯1) = 225900 J 5) Heating of 100.0 g of steam from 100.0 to 125.0 Celsius: (100.0 g) (25.0 K) (2.02 J g¯1 K¯1) = 5050 J 6) Summation of the results: 3090 + 33416 + 41840 + 225900 + 5050 = 309.3 kJ Problem 2: Determine the heat capacity of a coffee-cup calorimeter. During calibration 100.0 g of water at 58.5 °C has been added to 100.0 g of water, already in the calorimeter, at 22.8 °C. Calculate the heat capacity of the calorimeter in J/°C, if final temperature of the water is 39.7 °C. (Specific heat of water is 4.184 J/g °C.) Solution: 1) Heat given up by warm water: q = (100.0 g) (18.8 °C) (4.184 J/g °C) = 7865.92 J 2) Heat absorbed by water in the calorimeter: q = (100.0 g) (16.9 °C) (4.184 J/g °C) = 7070.96 J 3) The difference was absorbed by the calorimeter: 7865.92 - 7070.96 = 794.96 J 4) Calorimeter constant: 794.96 J / 16.9 °C = 47.0 J/°C Problem 3: Determine the final temperature when 10.0 g of aluminum at 130.0 °C mixes with 200.0 grams of water at 25.0 °C. Please note the starting temperature of the metal is above the boiling point of water. In reality, the sample may vaporize a tiny amount of water, but we will assume it does not for the purposes of the calculation. Solution: 1) The colder water will warm up and the warmer metal will cool down. The whole mixture will equilibrate up at the same temperature. The energy which "flowed" out of the warmer metal equals the energy which "flowed" into the colder water: Qaluminum = Qwater (10) (130 - x) (0.901) = (200.0 )(x - 25) (4.18) 117.13 - 0.901x = 83.6x - 2090 x = 26.12 °C. Important! Water didn’t cross temperature of phase transition – vaporization; otherwise calculation would be more complex. Calculation of reaction standard Gibbs free energy: For the general reaction aA + bB -> cC + dD ΔG°rxn = cΔGf°(C) + dΔGf°(D) - aΔGf°(A) - bΔGf°(B) Example: Calculate the Gibbs free energy for the following reaction at 25 °C. Cu (s) + H2O (g) -> CuO (s) + H2 (g) ΔG°rxn = ΔGf°(CuO (s)) – ΔGf°(H2O (g)) = (–129.7 kJ/mol) – (–228.6 kJ/mol) = 98.9 kJ/mol ΔGf° = 0; for elements in their standard state by definition. At equilibrium, ΔG = 0! Important points Application uses dot as a decimal separator. Special attention should be paid for units’ consistency.