The grams of aluminum that are required to produce 3.5 moles of AlO3 in presence of excess O2 is calculated as below
write the equation for reaction
4 Al + 3O2 =2 Al2O3
by use of mole ratio between Al to Al2O3 which is 4 :2 the moles of Al
=3.5 x4/2 = 7 moles
mass of Al = moles / x molar mass
= 7 moles x27 g/mol =189 grams
Rick Maurer i think that’s how you spell his last name
Answer: they giving you some hard question
Explanation:
i dont know know what the answer is do you have any answer options
Answer:
False
Explanation:
Atoms only achieve complete outer electron shells if they contain an outer shell with 7 electrons before gaining another electron or an outer shell with 1 electron before losing an electron. This is assuming that the octet-rule can be applied to said atom. In addition, the number of valence electrons varies from atom to atom which is why not ALL atoms achieve complete outer electron shells after gaining or losing just ONE electron.
Volume of Hydrogen V1 = 351mL
Temperature T1 = 20 = 20 + 273 = 293 K
Temperature T2 = 38 = 38 + 273 = 311 K
We have V1 x T2 = V2 x T1
So V2 = (V1 x T2) / T1 = (351 x 311) / 293 = 372.56
Volume at 38 C = 373 ml
