<span>2<span>C6</span><span>H6</span>O(l)+17<span>O2</span>−−>12C<span>O2</span>(g)+12<span>H2</span>O(l)</span>
<span>2S<span>O2</span>(g)+<span>O2</span>(g)−−>2S<span>O3</span>(g)</span>
<span><span>N2</span>(g)+<span>O2</span>(g)−−>2NO(g)</span>
<span>2Na(s)+B<span>r2</span>(l)−−>2NaBr(s)</span><span>
On the 1st 3 I have
12 -17 = -5
2 - 3 = -1
2 - 2 = 0
For the last one:
</span><span>Delta n=0</span><span>
</span>
<span>it must be a compound of a polyatomic cation (i think ammonium might be the only</span>
Answer:
HBr is a strong acid
Explanation:
KBr is a salt which makes a base . also KOH is a base
We have to get the relationship between metallic character and atomic radius.
Metallic character increases with increase in atomic radius and decrease with decrease of atomic radius.
If electrons from outermost shell of an element can be removed easily, that atom can be considered to have more metallic character.
With increase in atomic radius, nuclear force of attraction towards outermost shell electron decreases which facilitates the release of electron.
With decrease in atomic radius, nuclear force of attraction towards outermost shell electrons increases, so electrons are hold tightly to nucleus. Hence, removal of electron from outermost shell becomes difficult making the atom less metallic in nature.
Answer:
nBACO3=m/M=9,83/197=0,05(mol) ->nHCl=0,05.2/1=0,1(mol) =>VHCl=n/CM=0,1/0,44=0,227(lít)
Explanation:
