It will probably zip far from you and join itself to an adjacent molecule or atom. it gets to be distinctly radioactive when its core contains an excessive number of or an excessively couple of neutrons. Attempt to keep an indistinguishable number of neutrons and protons from you construct your iota. In the event that the awkwardness is excessively extraordinary, radioactive rot will happen.
<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>
They cause most of their damage when they reach land
132 g of C , 22 g of H , 176 g of O
132 + 22 + 176 => 330 g <span>of the substance
</span>Now convert the masses in <span>moles :
</span>
C = 12.0 u H = 1.0 u O = 16.0 u
C = 132 / 12.0 => 11 moles
H = 22 / 1.0 => 22 moles
O = 176 / 16.0 => 11 moles
Using the values obtained the lowest proportion in mols of elements present, simply divide the values found for the least of them<span>:
</span>
C = 11 / 11 => 1
H = 22 / 11 => 2
O = 11 / 11 => 1
formula empirically <span>is : CH</span>₂O
hope this helps!
The position of equilibrium lies far to the right, with products being favoured. Hence, option A is correct.
<h3>What is equilibrium?</h3>
Chemical equilibrium is a condition in the course of a reversible chemical reaction in which no net change in the amounts of reactants and products occurs.
A very high value of K indicates that at equilibrium most of the reactants are converted into products.
The equilibrium constant K is the ratio of the concentrations of products to the concentrations of reactants raised to appropriate stoichiometric coefficients.
When the value of the equilibrium constant is very high, the concentration of products is much higher than the concentration of reactants.
This means that most of the reactants are converted into products and the position of equilibrium lies far to the right, with products being favoured.
Hence, option A is correct.
Learn more about the equilibrium here:
brainly.com/question/23641529
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