<span>Forward & falling. Hope this helps!</span>
<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>
89.1% Au
The molar mass of Au2O3 is 2(196.97) + 3(16.00) = 441.94 g/mol
The mass percent of Au is therefore 2(196.97) / 441.94, which is 393.94/441.94 = 0.891 —> 89.1%
Answer:
False.
Explanation:
The given statement is false because for hot vacuum filtration, the filter paper should be wet rather than dry when pouring the hot solution into the Buchner funnel. This is because The possible explanation the filter paper needs to be wetted is not only to allow it to adhere to the funnel, but also to promote the solute to filter down readily across its pores of the paper without wetting it (this is true for organic and aqueous solvents).
The chemical breakdown of enormous quantities of organic material buried in the sedimentary rocks has produced ethane gas.
