Answer:
the mass required to inflate a 72 L bag is 191.491 g
Explanation:
reaction:
conditions:
- V = 72.0 L
- STP: P = 1 atm ∧ T = 298 K
gas law:
- PV = RTn
- R = 0.082 atm * L / K * mol
⇒ n = PV / RT
⇒ n = ((1 atm) * ( 72.0 L)) / (0.082 atm*L / K*mol) * (298 K)
⇒ n = 2.946 mol
⇒ m = n * Mw = ( 2.946 mol ) * ( 64.99 g/mol)
⇒ m = 191.491 g
<span>To find the mass of 3.00 moles of magnesium chloride (MgCl2), first record the atomic mass of magnesium (Mg) and chloride (Cl), which are both listed on the periodic table as follows:
Mg=24 g/mole
Cl=38 g/mole
Now, double the Cl mass since there are 2 Cl moles in MgCl2 and then add it to the Mg mass like so:
(38 g/mole*2 moles)+24 g/mole=100 g/mole
Finally, to calculate the mass of 3.00 moles of MgCl2, convert the combined atomic mass to grams as follows:
3.00 moles * 100 g/mole = 300 g</span>
mole=10 x 10⁻³ : 46 g/mol = 2.17 x 10⁻⁴
A. A pacemaker
Hope this helps!
