In a gas chromatography experiment, a mixture of two compounds (a and b) was separated for analysis. a chart recorder that fed paper at a constant rate of 100 mm. per minute was used to record the chromatogram, and a mark was made on the chart at the time the sample was injected. the first peak on the chromatogram corresponds to compound a, and that peak was observed at a distance of 5.83 cm from the sample injection point, and it had a height of 3.52 cm and a width at half-height of 1.25 cm. the peak corresponding to b was observed at a distance of 7.96 cm from the injection point, and it had a height of 3.34 cm and a width at half-height of 1.06 cm. what was the retention time for compound b (in seconds)? (enter your answer as a number only.)

47.76 seconds

Explanation:

Given:

Rate = 100 mm per minute

Distance of compound A from the Injection point = 5.83 cm = 58.3 mm

Distance of compound B from the Injection point = 7.96 cm = 79.6 mm

Now,

The retention time =

thus,

For the compound B

The retention time =

or

The retention time = 0.796 minute

also,

1 minute = 60 seconds

therefore,

0.796 minutes = 0.796 × 60 seconds = 47.76 seconds

given:

mass of na = 115 g

excess cl2

to determine:

mass of nacl produced

explanation:

given reaction is-

2na(s) + cl2(g) → 2nacl(s)

since cl2 is in excess, na will be the limiting reagent

as per the reaction stoichiometry na: nacl = 1: 1

i.e. moles of na reacted = moles of nacl formed

now, # moles of na = mass of na/atomic mass

                                  = 115 g/23 g.mol-1 = 5 moles

therefore, moles of nacl = 5

molar mass of nacl = 58 g/mol

mass of nacl = 5 moles * 58 g.mol-1 = 290 g

ans: amount of nacl produced = 290 g

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.