This indicates that the reaction is exothermic meaning that it releases heat/energy
The answer is 57.14%.
First we need to calculate molar mass of <span>NaHCO3. Molar mass is mass of 1 mole of a substance. It is the sum of relative atomic masses, which are masses of atoms of the elements.
Relative atomic mass of Na is 22.99 g
</span><span>Relative atomic mass of H is 1 g
</span><span>Relative atomic mass of C is 12.01 g
</span><span>Relative atomic mass of O is 16 g.
</span>
Molar mass of <span>NaHCO3 is:
22.99 g + 1 g + 12.01 g + 3 </span>· <span>16 g = 84 g
Now, mass of oxygen in </span><span>NaHCO3 is:
3 </span>· 16 g = 48 g
mass percent of oxygen in <span>NaHCO3:
48 g </span>÷ 84 g · 100% = 57.14%
Therefore, <span>the mass percent of oxygen in sodium bicarbonate is 57.14%.</span>
He realized that the physical and chemical properties of elements<span> were related to their atomic mass in a '</span>periodic<span>' way, and </span>arranged<span> them so that groups of </span>elements<span> with similar properties fell into vertical columns in </span>his table<span>.
</span><span>
</span>
21.4 g Al * (1 mol / 26.98 g ) * (2 mol Fe / 2 mol Al) = 0.793 mol Fe
91.3 g Fe2O3 * (1 mol / 159.69 g) * (2 mol Fe / 1 mol Fe2O3) = 1.14 mol Fe
0.793 mol Fe * (55.85 g / 1 mol) = 44.3 g Fe produced.
Xylene moles =\frac{17.12}{106.16×1000}=0.00016moles=
106.16×1000
17.12
=0.00016moles
Moles of CO_2 =\frac{56.77}{44.01×1000}=0.0013CO
2
=
44.01×1000
56.77
=0.0013
Moles of H_2O= =\frac{14.53}{18.02×1000}=0.0008H
2
O==
18.02×1000
14.53
=0.0008
Moles ratios
\frac{0.0013}{0.0008}=1.625
0.0008
0.0013
=1.625
\frac{0.0008}{0.0008}=1
0.0008
0.0008
=1
Hence molecular fomula
The empirical formula is C 4H 5.
The molecular formula C8H10
