Question

a 1.58 g sample of c2h3x3 (g) has a volume of 297 mL at 769 torr and 35 degree Celsius . identify element x

Answer 1

Do the same thing as before. Find the molar mass of the compound (C2H3X3) using the ideal gas law (PV=nRT) Subtract the mass of 2-carbons. Subtract the mass of 3-hydrogens. Divide the remaining mass by 3. On the periodic table, find the element with a mass equal to that number.

Answer 2

You need to convert the pressure into atm, and the temperature into kelvin or it doesn't work out mathematically

Answer 3

okay . got it thanks

Answer 4

I get X has a molar mass of 35.4g, this makes it chlorine. Your numbers may vary slightly

Answer 5

the answer is chlorine .

Answer 6

correct.

Answer 7

:)

Answer 8

can I ask you again bout the question before ?

Answer 9

in the 2nd part of the earlier question, the conditions of temp and pressure have changed, so the volume will be different. Plug in the moles you solved for earlier, and the NEW pressure and temp (in the proper units), and solve PV=nRT for VOLUME.

Answer 10

no , this one

if 200 cm3 of a gas of 0.268 g at STP .
a) calculate its molar mass

Answer 11

exact same question, except the volume is given in cm3, not mL.

By DEFINITION, 1cm3 = 1mL, so you can just substitute the numbers without changing them.

Answer 12

yes , but still . I did not get the right answer . the molar mass should be 30 but what do i get is a bit far

Answer 13

What did you plug in for a volume, and temperature? Temp has to be in kelvin, but you were given temperature in celsius. Volume has to be in liters, and pressure has to be in ATM

Answer 14

eh ? the temperature and volume are to be use at part b . not in part a

Answer 15

you still need to know the temp at STP, and you still need to plug in a volume to solve part a. It should look like:\[n = \frac{P*V}{R*T} = \frac{1atm * 0.200L}{0.08205\frac{L*atm}{mol*K}*273K} = 0.00893mol\]
Take the mass you're given, and divide by the moles you've solved for:
\[MM = \frac{g}{mol} = \frac{0.268g}{0.00893mol} = 30.0\frac{g}{mol}\]

Answer 16

okay . thanks . i'll try

Answer 17

can I know what is the volume that you get at part b ?