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일반화학5640

50.0 g metal 100.0℃ 24.8℃ 50.0 g water 51.2℃ 50.0 g metal 100.0℃ 24.8℃ 50.0 g water 51.2℃ Suppose 50.0 g of water in the Styrofoam cup has an in initial temperature of 24.8℃. When a 50.0 g piece of metal at a temperature of 100.0℃ is added, the temperature of the water and metal rise to final temperature of 51.2℃. What is the specific heat of this metal object? (Specific heat of water is 4.184 J/g•℃) 열용량과 비열 q = C m Δt ( 참고 https://ywpop.t.. 2024. 4. 20.
C3H6O3 0.100 g sample 0.0857 M NaOH 12.95 mL C3H6O3 0.100 g sample 0.0857 M NaOH 12.95 mL Lactic acid, C3H6O3, is the acid present in sour milk. A 0.100 g sample of pure lactic acid requires 12.95 mL of 0.0857 M sodium hydroxide for complete reaction. How many moles of hydroxide ion are required to neutralize one mole of lactic acid? C3H6O3의 몰질량 = 90.08 g/mol 이므로, 0.100 g / (90.08 g/mol) = 0.00111 mol C3H6O3 ( 참고 https://ywpop.tistory.com/.. 2024. 4. 20.
NaCl Na2SO4 NaNO3 mass Na 32.08% O 36.01% Cl 19.51% NaCl Na2SO4 NaNO3 mass Na 32.08% O 36.01% Cl 19.51% A sample containing NaCl, Na2SO4, and NaNO3 gives the following elemental analysis in percent by mass: Na = 32.08%, O = 36.01%, Cl = 19.51%. Calculate the percent by mass of each compound in the sample. 시료의 질량 = 100 g 이라 가정하면, 각 성분 원소의 질량 백분율(%) = 각 성분 원소의 질량(g) 각 성분 원소의 몰수를 계산하면, > Na의 몰수 = 32.08 g / (22.99 g/mol) = 1.395 mol ( 참고 https://ywpo.. 2024. 4. 20.
분자의 종류가 원자의 종류보다 훨씬 더 많은 이유 분자의 종류가 원자의 종류보다 훨씬 더 많은 이유 ▶ (분자를 구성하는 원자의 종류가 같아도) 결합하는 원자의 수나 결합 방식에 따라 다른 분자가 만들어지기 때문이다. ( 참고: 이성질체 https://ywpop.tistory.com/10881 ) ▶ (물질을 구성하는 원자의 종류가 같아도) 결합하는 원자의 수나 결합 방식에 따라 다른 물질이 만들어지기 때문이다. 가령, C, H, O, 이렇게 3개 원자만 있다고 가정하면, 이 3개 원자로 만들 수 있는 분자는 H2, O2, O3, H2O, H2O2, CH4, C6H6, CO, CO2, CH3OH, C2H5OH, ... ---> 거의 무한대 개의 분자를 만들 수 있다. 2024. 4. 20.
2H2O2(l) ΔH = -196 kJ H2O2(l) 5.00 g 2H2O2(l) ΔH = -196 kJ H2O2(l) 5.00 g Hydrogen peroxide can decompose to water and oxygen by the reaction: 2H2O2(l) → 2H2O(l) + O2(g) ... ΔH = –196 kJ Calculate the quantity of heat released when 5.00 g of H2O2(l) decomposes at constant pressure. 2H2O2(l) → 2H2O(l) + O2(g) ... ΔH = –196 kJ 2몰 H2O2의 분해열이 –196 kJ 이므로, ΔH = –196 kJ / 2 mol = –98 kJ/mol ---> H2O2 1몰당 분해열 H2O2의 몰질량 = 34.01 g/mol 이므로, .. 2024. 4. 20.
one gallon 2.0 g lead 10000. miles 25 miles per gallon one gallon 2.0 g lead 10000. miles 25 miles per gallon If one gallon of gasoline contains about 2.0 g of lead, how much lead (in kg) is emitted into the atmosphere from a car that travels 10,000. miles a year and averages 25 miles per gallon of gasoline? [참고] 환산 인자 [ https://ywpop.tistory.com/3121 ] (10000. mi) (1 gal / 25 mi) (2.0 g Pb / 1 gal) = (10000.) (1 / 25) (2.0) = 800 g Pb = 0.80 kg Pb .. 2024. 4. 20.
lead tin 3.50 g HNO3 1.57 g PbSO4(s) lead tin 3.50 g HNO3 1.57 g PbSO4(s) A 3.50 g ofan alloy which contains only lead and tin is dissolved in hot HNO3. Excess sulfuric acid is added to this solution and 1.57 g of PbSO4(s) is obtained. a) Write the net ionic equation for the formation of PbSO4(s). b) Assuming all the lead in the alloy reacted to form PbSO4, what was the amount, in grams, of lead and tin in the alloy respectively? n.. 2024. 4. 20.
yield 67.0% N2H4 mass 26.57 g N2 react 4.65 g H2 yield 67.0% N2H4 mass 26.57 g N2 react 4.65 g H2 For the reaction N2(g) + 2H2(g) → N2H4(l), if the percent yield for this reaction is 67.0%, what is the actual mass of hydrazine (N2H4) produced when 26.57 g of nitrogen reacts with 4.65 g of hydrogen? 각 반응물의 몰수를 계산하면, 26.57 g / (28 g/mol) = 0.9489 mol N2 ( 참고 https://ywpop.tistory.com/7738 ) 4.65 g / (2 g/mol) = 2.325 mol H2 N2(g) + 2H2(g) → N2H4.. 2024. 4. 19.
27℃ 10.0 L O2 50.0 g He 300 g total pressure (atm) 27℃ 10.0 L O2 50.0 g He 300 g total pressure (atm) 27℃, 10.0 L 통에 산소 50.0 g과 헬륨 300 g이 들어있다. 이 통에 들어있는 기체의 전체 압력(atm)은? At 27℃, a 10.0 L container contains 50.0 g of oxygen and 300 g of helium. What is the total pressure (atm) of the gas in this container? > O2의 몰질량 = 32.00 g/mol > He의 몰질량 = 4.00 g/mol > n = W / M ( 참고 https://ywpop.tistory.com/7738 ) PV = nRT 로부터, ( 참고 https://ywpop.tistory.com.. 2024. 4. 19.
1.00 atm 295 K 2.25 L CO2 C6H12O6 1.00 atm 295 K 2.25 L CO2 C6H12O6 Yeast and other organisms can convert glucose (C6H12O6) to ethanol (CH3CH2OH) by a process called alcoholic fermentation. The net reaction is C6H12O6(s) → 2C2H5OH(l) + 2CO2(g) Calculate the mass of glucose required to produce 2.25 L of CO2 measured at P = 1.00 atm and T = 295 K. PV = nRT 로부터, ( 참고 https://ywpop.tistory.com/3097 ) CO2의 몰수를 계산하면, n = PV / RT = [(1.. 2024. 4. 19.
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