The tonga trench in the pacific ocean is 36,000 feet deep. assuming that sea water has an average density of 1.04 g/cm3, calculate the absolute (total) pressure at the bottom of the trench in atmospheres. (1.00 in = 2.54 cm, 1.00 atm = 1.01 * 105 pa)

1110 atm

Let's start by calculating how many cm deep is 36,000 feet.

36000 ft * 12 in/ft * 2.54 cm/in = 1097280 cm

Now calculate how much a column of water 1 cm square and that tall would mass.

1097280 cm * 1.04 g/cm^3 = 1141171.2 g/cm^2

We now have a number using g/cm^2 as it's unit and we desire a unit of Pascals (kg / (m*s^2)).

It's pretty obvious how to convert from g to kg. But going from cm^2 to m is problematical. Additionally, the s^2 value is also a problem since nothing in the value has seconds as an unit. This indicates that a value has been omitted. We need something with a s^2 term and an additional length term. And what pops into mind is gravitational acceleration which is m/s^2. So let's multiply that in after getting that cm^2 term into m^2 and the g term into kg.

1141171.2 g/cm^2 / 1000 g/kg * 100 cm/m * 100 cm/m = 11411712 kg/m^2

11411712 kg/m^2 * 9.8 m/s^2 = 111834777.6 kg / (m*s^2) = 111834777.6 Pascals

Now to convert to atm

111834777.6 Pa / 1.01x10^5 Pa/atm = 1107.2750 atm

Now we gotta add in the 1 atm that the atmosphere actually provides (but if you look closely, you'll realize that it won't affect the final result).

1107.274 atm + 1 atm = 1108.274 atm

And finally, round to 3 significant figures since that's the accuracy of our data, giving 1110 atm.