<h2>
</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 -
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. Formation of ammonia
- The formation of ammonia is by reacting nitrogen gas and hydrogen gas.
→ 
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.
Answer:
1) Fe = 69.9%
O = 31.1%
2) H = 5.19%
O = 16.5%
N = 28.9%
C = 49.5%
Explanation:
One easy way to do percent compositions is to assume you have 100g of a substance.
1) Lets say we have 100g of Fe2O3.
The total molar mass would be:
The molar mass of the Fe2 alone is:
Thus, the grams of Fe2(out of a 100) could be calculated by multiplying 100g * the molar mass ratio of Fe2 to the whole:
Which is approximately 69.9%.
We can find the amount of O3 by simply subtracting, as the rest of the compound is made of O3. Thus, the % composition of O3 is 31.1%
You can then do this same process to the next question, getting us the following:
H = 5.19%
O = 16.5%
N = 28.9%
C = 49.5%
It changes precipitation patterns, causes strong storms, and transfers warm and cold air.
The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).
ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
<span> ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 </span>°C<span>
</span><span> ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 </span>°C<span>
</span><span> ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 </span>°C<span>
</span><span> ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 </span>°C<span>
</span>
Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.
