If the He H e injected into the abdomen produces a pressure of 22 mmHg m m H g and a volume of 2.7 L L at 31 ∘C ∘ C , how many grams of He H e were used?

1280.27 g

Explanation:

Pressure = 22 mmHg / 760 = 0.0289 atm (Converting to atm by dividing by 760)

Volume = 2.7 L

Temperature = 31 ∘C + 273 = 304 K

Mass = ?

Using the ideal gas equation;

PV = nRT

n = RT / PV = 0.0821 * 304 / ( 0.0289 * 2.7)

n = 319.86 mol

Mass = Number of moles * Molar mass

Mass = 319.86  * 4.002602

Mass = 1280.27 g

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.