How would I calculate where the buried inspection chamber is from the details below?

Posted on January 9th, 2010 by admin in chamber | No Comments »

Also -The co-ordinates of the inspection chamber.
– WCB and distance from station A

Buried
Chamber SW2

43.525m distance from chamber SW1 to SW2
27.810m distance from chamber SW3 to SW2

Inspection chamber SW1
140E
IL 42.300
CL 45.500

BC2
Inspection chamber SW3
191.625E
263.350N
IL 42.500
CL 46.000

Station A
120E, 260N
45,500

i’m not familiar with the terms IL and CL, here.

the E and N pairs look like latitude+longitude but usually N only goes as far as 90 degrees, if that were the case; so you want to find Cartesian co-ordinates with equal x-y scales, for as many of these as you can; then you can draw a to-scale map, showing points you know, and estimating those you don’t know.

extend circles around the points SW1 and SW3 with radii 43.525 and 27.810 respectively, making adjustments until things fit.

algebraically, it becomes a system of polynomial equations where circles intersect, so you might have expressions of the form (x-a)² + (y-b)² = R² to map each of the circles

How many grams of Oxygen are required to fill a 3.00 x 10^3 liter hyperbaric chamber at a pressure of 300cm Hg?

Posted on January 2nd, 2010 by admin in hyperbaric chamber | 2 Comments »

and at 75F?

75 F – 32 = 1.8 Celsius
43 = 1.8 Celsius
temp = 23.9 Celsius ==== 297 Kelvin

300 cm Hg @ 1 atm / 76 cm Hg = 3.947 atm

find moles of oxygen
PV = nRT
(3.947 atm)(3000 litres) = n (0.0821 L-atm/mol-K)(297Kelvin)
n = 485.66 moles of oxygen

use molar mass to find mass of oxygen:
485.66 moles oxygen @ 32.00 g/mol =
your answer:
15,541 grams of oxygen

but if this is a 3 sigfig problem, we should round off to
15,500 grams of oxygen

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