반응형 화학/원소(연소)분석308 85.62% C molar mass 84.16 g/mol molecular formula 85.62% C molar mass 84.16 g/mol molecular formula C와 H로만 이루어진 화합물에서 85.62%의 C가 발견되었다. 분자량이 84.16 g/mol이라면 화합물의 분자식을 결정하시오. A compound containing only carbon and hydrogen was found to be 85.62% carbon. Determine the molecular formula of the compound if its molar mass is 84.16 g/mol. --------------------------------------------------- 답: C6H12 [Blogger 블로그스팟] [ https://ywpop.blogspot.com/2023/10/8.. 2023. 10. 31. 2.500 g sample CHO 3.689 g CO2 1.500 g H2O 2.500 g sample CHO 3.689 g CO2 1.500 g H2O 2.500 g sample containing C, H, O produces 3.689 g CO2, 1.500 g H2O --------------------------------------------------- [1] 1.007 g C [2] 0.168 g H [3] 1.325 g O [Blogger 블로그스팟] [ https://ywpop.blogspot.com/2023/10/2500-g-sample-cho-3689-g-co2-1500-g-h2o.html ] [키워드] CHO 연소분석 기준, CHO 원소분석 기준, CHO 연소분석 사전, CHO 원소분석 사전 M2(CO3)3 화합물 속 금속 Which metal (ion) .. 2023. 10. 9. 원소분석. X Cl mass 13.10% X six times 원소분석. X Cl mass 13.10% X six times You find a compound composed only of element X and chlorine, and you know that the compound is 13.10% X by mass. Each molecule of the compound contains six times as many chlorine atoms as X atoms. What is element X? --------------------------------------------------- ▶ 참고: 원소분석 [ https://ywpop.tistory.com/64 ] ---------------------------------------------------.. 2023. 7. 3. 4.07% H, 24.27% C, 71.65% Cl mol mass 98.96 g 4.07% H, 24.27% C, 71.65% Cl mol mass 98.96 g A compound contains 4.07% H, 24.27% C and 71.65% Cl. Its molar mass is 98.96 g. What are its empirical and molecular formulas? --------------------------------------------------- ▶ 참고: 원소 분석 [ https://ywpop.tistory.com/64 ] --------------------------------------------------- 각 성분의 몰수 계산 > 24.27 g / (12.01 g/mol) = 2.02 mol C > 4.07 / 1.008 = 4.04 mol.. 2023. 5. 8. 질량 기준 25% A, 75% B이고 B 원자량은 A의 2배 질량 기준 25% A, 75% B이고 B 원자량은 A의 2배 질량 기준 25% A, 75% B이고, B의 원자량은 A의 2배인 화합물의 실험식 --------------------------------------------------- ▶ 참고: 원소분석 [ https://ywpop.tistory.com/64 ] --------------------------------------------------- A의 원자량 = 1 g/mol 이라 두면, B의 원자량 = 2 g/mol. 각 성분의 몰수 계산 > A의 몰수 = 25 g / (1 g/mol) = 25 mol > B의 몰수 = 75 g / (2 g/mol) = 37.5 mol 몰수의 가장 작은 정수비 계산 A : B = 25 : 37.5 = 25/25.. 2023. 5. 6. 85.5% C, 14.2% H gas 4 g 27℃ 1 atm 3.5 L 85.5% C, 14.2% H gas 4 g 27℃ 1 atm 3.5 L 85.5% 탄소와 14.2% 수소로 된 어떤 기체 화합물 4 g이 27℃ 1기압에서 3.5 L의 부피를 차지한다. 기체의 분자식은? --------------------------------------------------- ▶ 참고: 원소분석 [ https://ywpop.tistory.com/64 ] --------------------------------------------------- 각 성분의 몰수 계산 > 85.5 g / (12 g/mol) = 7.125 mol C > 14.2 / 1 = 14.2 mol H ( 참고 https://ywpop.tistory.com/7738 ) 몰수의 가장 작은 정수비 계산 C : H = .. 2023. 5. 3. C H N sample 5.250 mg CO2 14.242 mg H2O 4.083 mg C H N sample 5.250 mg CO2 14.242 mg H2O 4.083 mg Nicotine, a component of tobacco, is composed of C, H, and N. A 5.250 mg sample of nicotine was combusted, producing 14.242 mg of CO2 and 4.083 mg of H2O. What is the empirical formula for nicotine? If nicotine has a molar mass of 160±5 g/mol, what is its molecular formula? --------------------------------------------------- ▶ 참고: 원소분석 [ https://y.. 2023. 4. 21. 0.240 g CaCO3 0.374 g BaSO4 0.206 g NH3 FW 156 0.240 g CaCO3 0.374 g BaSO4 0.206 g NH3 FW 156 A compound containing Ca, C, N and S was subjected to quantitative analysis and formula mass determination. A 0.375 g of this compound was mixed with Na2CO3 to convert all Ca into 0.240 g CaCO3. A 0.125 g sample of compound was carried through a series of reactions until all its S was changed into SO4^2- and precipitated as 0.374 g of BaSO4. A 0.946.. 2023. 4. 17. atenolol CHON 5.000 g 11.57 g CO2 3.721 g H2O 10.52% N atenolol CHON 5.000 g 11.57 g CO2 3.721 g H2O 10.52% N Beta-blockers are a class of drug widely used to manage hypertension. Atenolol, a beta-blocker, is made up of carbon, hydrogen, oxygen, and nitrogen atoms. When a 5.000 g sample is burned in oxygen, 11.57 g of CO2 and 3.721 g of H2O are obtained. A separate experiment using the same mass of sample (5.000 g) shows that atenolol has 10.52% nit.. 2023. 4. 8. C3H4O2 0.275 g H2O 0.102 g CO2 0.374 g C3H4O2 0.275 g H2O 0.102 g CO2 0.374 g Propenoic acid, C3H4O2, is a reactive organic liquid that is used in the manufacturing of plastics, coatings, and adhesives. An unlabeled container is thought to contain this liquid. A 0.275 g sample of the liquid is combusted to produce 0.102 g of water and 0.374 g carbon dioxide. Is the unknown liquid propenoic acid? Support your reasoning with calculatio.. 2023. 4. 4. 이전 1 2 3 4 5 ··· 31 다음 반응형
