Melted rock and minerals below the earth's crust is also known as magma
<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>
Answer:
π = 14.824 atm
Explanation:
wt % = ( w NaCL / w sea water ) * 100 = 3.5 %
assuming w sea water = 100 g = 0.1 Kg
⇒ w NaCl = 3.5 g
osmotic pressure ( π ):
∴ T = 20 °C + 273 = 293 K
∴ C ≡ mol/L
∴ density sea water = 1.03 Kg/L....from literature
⇒ volume sea water = 0.1 Kg * ( L / 1.03 Kg ) = 0.097 L sln
⇒ mol NaCl = 3.5 g NaCL * ( mol NaCL / 58.44 g ) = 0.06 mol
⇒ C NaCl = 0.06 mol / 0.097 L = 0.617 M
⇒ π = 0.617 mol/L * 0.082 atm L / K mol * 293 K
⇒ π = 14.824 atm
