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>
Answer:
18.65004 grams H2O
Explanation:
First, we need to write down the balanced chemical equation for the decomposition reaction:
2LiOH -> H2O + Li2O
Since we have grams of LiOH and we need to know the grams of water, we need to convert to moles since we can only compare moles to moles.
The amu of LiOH is 23.947.
The given grams of LiOH is 63.. To convert to moles, we will divide 63 by 23.947..
This gives us 2.6310 moles LiOH..
To convert to moles of H2O (and later grams of H2O), we will use the mole fractions from the balanced equation...
When we look at the balanced equation we can see that 2 moles of LIOH can produce 1 mol of Water, so:
2.6310 moles
= 1.3155 moles H2O
Now we will convert from moles to grams (we must multiply by the amu)
1.3155 moles H2O = 18.65 grams H2O
Answer:
The atom that loses the electrons becomes a positively charged ion, while the one that gains them becomes a negatively charged ion
Explanation:
here's the molecule you were looking for