Key Question #3

# Key Question #3 - 1 mol = 4.97g 0.01838mol = 270.4 g/mol...

This preview shows pages 1–2. Sign up to view the full content.

Lesson 3 Key Questions 13) Molar mass of TiO 2 = 79.88 g/mol Molar mass of AgCl = 143.32 g/mol 1.2g / 79.88g/mol = 0.015 mol of Ti atoms 6.45g / 1443.32 = 0.045 mol of Cl atoms n= 0.045mol / 0.015mol n=3 The ratio of Ti to Cl is 1:3 The simplest formula of the original compound is TiCl 3 14) Moles of C = 80.0g / 12.0g/mol = 6.67 mol Moles of H = 8.2g / 1.01g/mol = 8.12 mol Moles of O = 11.8g / 16g/mol = 0.75 mol C = 6.67mol / 0.75mol = 9 H = 8.12mol / 0.75mol = 11 O = 0.75mol / 0.75mol = 1 The empirical formula C 9 H 11 O n=(PV)/(RT) n=(0.9atm x 0.793L) / (0.0821 kPa L x 473K) n=0.01838 mol 0.01838 mole of Estrone has a mass of 4.97g

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 mol = 4.97g / 0.01838mol = 270.4 g/mol molar mass of C 9 H 11 O = 135.2 g/mol n= 270.4 g/mol / 135.2g/mol n=2 The molecular formula for Estrone is C 18 H 22 O 2 15) Ammonia = NH 3 Ammonium nitrate = NH 4 NO 3 Molar mass = 17 g/mol Molar mass = 80.1 g/mol % of N = 14.01 g/mol / 17g/mol % of N = 28g/mol / 80.1g/mol % of N = 82.9 % of N = 35% The liquid ammonia (NH 3 ) would be better because it has a higher concentration of Nitrogen by mass. The ammonia has 82.9% Nitrogen compare to 35% Nitrogen of the ammonium nitrate....
View Full Document

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern