Answer:
B
Explanation:
"I think my bike is making a scraping noise because the bearings are not lubricated."
Answer:
.
Explanation:
Electrons are conserved in a chemical equation.
The superscript of
indicates that each of these ions carries a charge of
. That corresponds to the shortage of one electron for each
ion.
Similarly, the superscript
on each
ion indicates a shortage of three electrons per such ion.
Assume that the coefficient of
(among the reactants) is
, and that the coefficient of
(among the reactants) is
.
.
There would thus be
silver (
) atoms and
aluminum (
) atoms on either side of the equation. Hence, the coefficient for
and
would be
and
, respectively.
.
The
ions on the left-hand side of the equation would correspond to the shortage of
electrons. On the other hand, the
ions on the right-hand side of this equation would correspond to the shortage of
electrons.
Just like atoms, electrons are also conserved in a chemical reaction. Therefore, if the left-hand side has a shortage of
electrons, the right-hand side should also be
electrons short of being neutral. On the other hand, it is already shown that the right-hand side would have a shortage of
electrons. These two expressions should have the same value. Therefore,
.
The smallest integer
and
that could satisfy this relation are
and
. The equation becomes:
.
Answer:
a) Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)
b) 3.14g must be added
Explanation:
a) For the reaction:
Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) + H₂O(l) + NH₃(g)
As you see, there are 8 moles of water in reactants and 2 moles of oxygen in octahydrate, thus, water moles must be 10:
Ba(OH)₂.8H₂O(s) + NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + NH₃(g)
To balance hydrogens, the other coefficients are:
Ba(OH)₂.8H₂O(s) + <em>2 </em>NH₄SCN(s) → Ba(SCN)₂(s) +<em>10</em> H₂O(l) + <em>2</em> NH₃(g)
b) As you see in the balanced reaction, 1 mole of barium hydroxide octahydrate reacts with 2 moles of NH₄SCN. 6.5g of Ba(OH)₂.8H₂O are:
6.5 g × (1mol / 315.48g) =<em> 0.0206moles of Ba(OH)₂.8H₂O</em>. Thus, moles of NH₄SCN that must be used for a complete reaction are:
0.0206moles of Ba(OH)₂.8H₂O × ( 2 mol NH₄SCN / 1 mol Ba(OH)₂.8H₂O) = <em>0.0412moles of NH₄SCN</em>. In grams:
0.0412moles of NH₄SCN × ( 76.12g / 1mol) = <em>3.14g must be added</em>