You have too many s.f. in that answer for A. Also, I don't get that exactly; the problem, I suspect, is that you rounded with intermediate numbers and I didn't.
B is ok.
6.3 mg of a boron hydride is contained in a flask of 385 mL at 25. degrees C and a pressure of 11 Torr.
A. determine the molar mass of the hydride.
B. which of the following hydrides is contained in the flask, BH3, B2H6, or B4H10
A. MM = 27.597 g/mol
B. B2H6
4 answers
Here is my work, I might have gotten the answers the wrong way.
6.3 mg / 1000 = .0063g
385 mL / 1000 = .385 L
25 degrees C + 273 = 298 K
11 Torr / 760 = .0145 atm
.0145 atm x .385L = n (.08206 Latm/molK) 298K
n = 2.28x10^-4 moles
A. MM = .0063g / 2.28x10^-4mole = 27.597 g/mol
B. BH3 molar mass = 13.83 g
B2H6 = 27.67 g - this one is closest...
B4H10 = 53.32 g
6.3 mg / 1000 = .0063g
385 mL / 1000 = .385 L
25 degrees C + 273 = 298 K
11 Torr / 760 = .0145 atm
.0145 atm x .385L = n (.08206 Latm/molK) 298K
n = 2.28x10^-4 moles
A. MM = .0063g / 2.28x10^-4mole = 27.597 g/mol
B. BH3 molar mass = 13.83 g
B2H6 = 27.67 g - this one is closest...
B4H10 = 53.32 g
You did it right; I used 11/760 and left it in the calculator.
A volume of 26.5 mL of nitrogen gas was collected in a tube at a temperature of 17° C and a pressure of 737 mm Hg. The next day the volume of the nitrogen was 27.1 mL with the barometer still reading 737 mm Hg. What was the temperature on the second day?
0 answers