<span>4 Al + 3 O2 → 2 Al2O3
(10.0 g Al) / (26.98154 g Al/mol) = 0.37062 mol Al
(19.0 g O2) / (31.99886 g O2/mol) = 0.59377 mol O2
0.37062 mole of Al would react completely with 0.37062 x (3/4) = 0.277965 mole of O2, but there is more O2 present than that, so O2 is in excess.
((0.59377 mol O2 initially) - (0.277965 mol O2 reacted)) x (31.99886 g O2/mol) =
10.1 g O2 left over</span><span>
</span>
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Answer:
0.125 moles
Explanation:
2.8 litres is equivalent to 2.8dm³
At STP,
1 mole = 22.4 dm³
x mole = 2.8 dm³
Cross multiply
22.4x = 2.8
Divide both sides by 22.4
x = 2.8/22.4
x = 0.125
Atomic mass of helium is 4.002642g/mol
(542000g)/(4.002642g/mol)*6.02*10^23 = 8.15*10^28 atoms
Answer:
cesium
In particular, cesium (Cs) can give up its valence electron more easily than can lithium (Li). In fact, for the alkali metals (the elements in Group 1), the ease of giving up an electron varies as follows: Cs > Rb > K > Na > Li with Cs the most likely, and Li the least likely, to lose an electron
Explanation:
