Combustion analysis of a hydrocarbon produced 0.40g of CO2 and 0.205g of H20. Calculate the
empirical formula of the hydrocarbon.

Convert grams of CO2 to moles of CO2 and then use the mole ratio of 1 mol C / 1 mol CO2 (since every mole of CO2 has 1 mol of carbon in it) to find moles of carbon.

Do the same for H2O except the mole ratio is 2 mol H / 1 mol H2O, since every mole of H2O has two moles of hydrogen in it.

0.40 g CO2 • 1 mol CO2/44.01 g CO2 • 1 mol C / 1 mol CO2 = 0.0091 mol C

0.205 g H2O • 1 mol H2O/18.02 g H2O • 2 mol H / 1 mol H2O = 0.0228 mol H

Now divide by the smallest number of moles (0.0091 moles of C) to get whole number values for both C and H rather than decimals.

0.0091/0.0091 = 1 mol C
0.0228/0.0091 = 2.5 mol H

You need whole numbers for both C and H, which you can get by just multiplying both by 2 (since 2.5 • 2 is 5).

So, your empirical formula ends up being C2H5.

Work done = force ×distancebut force = mass ×gravityand gravity = 10m\s force = 2×10 = 20 work done = 20×3 = 60 joules

Idk the answer sorry