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>
Random or ionic bond pattern
Answer:
A cuz a heterogeneous mixture is no uniform
<h2>
![H_2O + H_2 + O_2](https://tex.z-dn.net/?f=H_2O%20%20%2B%20H_2%20%2B%20O_2)
</h2>
Explanation:
1. Water decomposition
- Decomposition reactions are represented by-
The general equation: AB → A + B.
- Various methods used in the decomposition of water are -
- Electrolysis
- Photoelectrochemical water splitting
- Thermal decomposition of water
- Photocatalytic water splitting
- Water decomposition is the chemical reaction in which water is broken down giving oxygen and hydrogen.
- The chemical equation will be -
![H_2O + H_2 + O_2](https://tex.z-dn.net/?f=H_2O%20%20%2B%20H_2%20%2B%20O_2)
Hence, balancing the equation we need to add a coefficient of 2 in front of
on right-hand-side of the equation and 2 in front of
on left-hand-side of the equation.
∴The balanced equation is -
→ ![2 H_2 + O_2](https://tex.z-dn.net/?f=2%20H_2%20%2B%20O_2)
2. Formation of ammonia
- The formation of ammonia is by reacting nitrogen gas and hydrogen gas.
→ ![NH_3](https://tex.z-dn.net/?f=NH_3)
Hence, for balancing equation we need to add a coefficient of 3 in front of hydrogen and 2 in front of ammonia.
∴The balanced chemical equation for the formation of ammonia gas is as follows -
→
.
- When 6 moles of
react with 6 moles of
4 moles of ammonia are produced.