This is a reaction of combustion of methane
CH₄ + 2O₂ = CO₂ + 2H₂O
k(H₂O)=2
You’ve not put anything below so I’m not sure what the options are but the relative charge of a proton is +1
<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>
Answer:
true
Explanation:
what is the answer? please
