Three glass bulbs, joined by closed stopcocks, have the following volumes and initial pressures of the specified gases.
bulb a: 200. ml of kr(g) at 190. torr
bulb b: 400. ml of h2s(g) at 1.00 atm
bulb c: 1.00 l of n2(g) at 75.994 kpa

after both stopcocks are opened and the gases allowed to diffuse throughout, what will be the ultimate total pressure?

P = 0.75 atm

Explanation:

Using Boyle's law  

Given ,  

For Bulb A

Pressure = 190 torr

The conversion of P(torr) to P(atm) is shown below:

So,  

Pressure = 190 / 760 atm = 0.25 atm

Volume = 200 mL

For Bulb B

Pressure = 1 atm

Volume = 400 mL

For Bulb C

Volume = 1.00 L = 1000 mL

Pressure = 75.994 kPa

The expression for the conversion of pressure in KiloPascal to pressure in atm is shown below:

P (kPa) = P (atm)

75.994 kPa = atm

Pressure = 0.75 atm

Also, Total volume, V = 200 + 400 + 1000 mL =  1600 mL

Using above equation as:

We get, Total pressure, P = 0.75 atm

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