반응형 일반화학/[03장] 화학량론791 6.78×10^22 개의 산소 원자를 포함하는 H2SO4의 몰수 6.78×10^22 개의 산소 원자를 포함하는 H2SO4의 몰수 moles of H2SO4 containing 6.78×10^22 oxygen atoms (6.78×10^22 개) (1 mol / 6.022×10^23 개) = (6.78×10^22) (1 / (6.022×10^23)) = 0.113 mol O ( 참고 https://ywpop.tistory.com/6475 ) 개수비 = 몰수비 이고, 1개 H2SO4 속에 들어있는 O의 개수 = 4 이므로, H2SO4 : O = 1 : 4 = ? mol : 0.113 mol ? = 0.113 / 4 = 0.0283 mol H2SO4 답: 0.0283 mol [키워드] 분자 속 원자 기준, 분자에 들어있는 원자 기준 2024. 4. 21. 0.0946 M KMnO4 26.2 mL Ca^2+ 24 hour 100 300 mg 0.0946 M KMnO4 26.2 mL Ca^2+ 24 hour 100 300 mg Calcium in blood or urine can be determined by precipitation as calcium oxalate, CaC2O4. The precipitate is dissolved in strong acid and titrated with potassium permanganate. The products of the reaction are carbon dioxide and manganese(II) ion. A 24 hour urine sample is collected from an adult patient, reduced to a small volume, and titrated with .. 2024. 4. 21. 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. 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. 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. Calicheamicin C55H74IN3O21S4 5.0×10^(-9) M 25.00 mL Calicheamicin C55H74IN3O21S4 5.0×10^(-9) M 25.00 mL Calicheamicin gamma-1, C55H74IN3O21S4, is one of the most potent antibiotics known: one molecule kills one bacterial cell. Describe how you would (carefully!) prepare 25.00 mL of an aqueous calicheamicin gamma-1 solution that could kill 1.0×10^8 bacteria, starting from a 5.00×10^(-9) M stock solution of the antibiotic. 칼리키아마이신 감마-1(Calicheamici.. 2024. 4. 14. 0.300 M NaOH 105.0 mL 0.060 M AlCl3 150.0 mL 0.300 M NaOH 105.0 mL 0.060 M AlCl3 150.0 mL A solution of 105.0 mL of 0.300 M NaOH is mixed with a solution of 150.0 mL of 0.060 M AICI3. a) Write the balanced chemical equation for the reaction that occurs. b) What precipitate forms? c) What is the limiting reactant? d) How many grams of this precipitate form? e) What is the concentration of each ion that remains in solution? a) Write the bala.. 2024. 4. 13. 2.50 mol Ba(OH)2의 질량 2.50 mol Ba(OH)2의 질량 ① 물질의 몰질량을 계산한다.( 참고 https://ywpop.tistory.com/16840 )( 참고 https://ywpop.tistory.com/18326 ) ② 몰수를 질량으로 환산한다.( 참고: n=W/M https://ywpop.tistory.com/7738 ) ① Ba(OH)2의 몰질량 = 171.34 g/mol 이므로,② 2.50 mol × (171.34 g/mol) = 428.35 g 답: 428 g [ 관련 글 https://ywpop.tistory.com/24730 ]10.0 g KHCO3의 몰수 [ 관련 글 https://ywpop.blogspot.com/2024/05/ca-27.. 2024. 4. 11. SO2의 조성 백분율 SO2의 조성 백분율 > SO2의 몰질량 = (32) + 2(16) = 64 g/mol ( 참고 https://ywpop.tistory.com/16840 ) > S의 조성 백분율 = [32 / 64] × 100 = 50% S ( 참고 https://ywpop.tistory.com/12958 ) > O의 조성 백분율 = [2(16) / 64] × 100 = 50% O [ 관련 글 https://ywpop.tistory.com/5171 ] CO2의 조성 백분율 [키워드] SO2의 조성 백분율 기준 2024. 4. 11. 이전 1 2 3 4 5 ··· 80 다음 반응형