Molar mass:
O2 = 16 x 2 = 32.0 g/mol Mg = 24 g/mol
<span>2 Mg(s) + O2(g) --->2 MgO(s)
</span>
2 x 24.0 g Mg -------------> 32 g O2
5.00 g Mg -----------------> ( mass of O2)
mass of O2 = 5.00 x 32 / 2 x 24.0
mass of O2 = 160 / 48
= 3.33 g of O2
hope this helps!
Since Au is a symbol for Gold, and once you split the name into to giving each ion its charge... you'll see that this compound has Au+2 and Cl03- .... so the name would be
Gold(II) Chlorate
Hope this helps!
Answer : The limiting reagent is 
Solution : Given,
Moles of methane = 2.8 moles
Moles of = 5 moles
Now we have to calculate the limiting and excess reagent.
The balanced chemical reaction is,
From the balanced reaction we conclude that
As, 2 mole of react with 1 mole of 
So, 5 moles of react with 
moles of
From this we conclude that, is an excess reagent because the given moles are greater than the required moles and is a limiting reagent and it limits the formation of product.
Hence, the limiting reagent is
So it would be the complimentary base pairing, meaning that the codon must have been:
GAC
(Which is the codon for aspartic acid)
