Answer:
9.63 L of NO
Explanation:
We'll begin by calculating the number of mole in 50.0 g of NH₄ClO₄. This can be obtained as follow:
Mass of NH₄ClO₄ = 50 g
Molar mass of NH₄ClO₄ = 14 + (4×1) + 35.5 + (16×4)
= 14 + 4 + 35.5 + 64
= 117.5 g/mol
Mole of NH₄ClO₄ =?
Mole = mass /molar mass
Mole of NH₄ClO₄ = 50/117.5
Mole of NH₄ClO₄ = 0.43 mole
Next, we shall determine the number of mole of NO produced by the reaction of 50 g (i.e 0.43 mole) of NH₄ClO₄. This can be obtained as follow:
3Al + 3NH₄ClO₄ –> Al₂O₃ + AlCl₃ + 3NO + 6H₂O
From the balanced equation above,
3 moles of NH₄ClO₄ reacted to produce 3 moles of NO.
Therefore, 0.43 mole of NH₄ClO₄ will also react to produce 0.43 mole of NO.
Finally, we shall determine the volume occupied by 0.43 mole of NO. This can be obtained as follow:
1 mole of NO = 22.4 L
Therefore,
0.43 mole of NO = 0.43 × 22.4
0.43 mole of NO = 9.63 L
Thus, 9.63 L of NO were obtained from the reaction.
Because more than one substance was released (following a color change signifying a chemical reaction), the sample was indeed, a compound.
![pH=-\log_{10} [H_3O^+] \Rightarrow [H_3O^+]=10^{-pH} \\ \\ pH=2.8 \\ \ [H_3O^+]=10^{-2.8} \approx 1.58 \times 10^{-3}](https://tex.z-dn.net/?f=pH%3D-%5Clog_%7B10%7D%20%5BH_3O%5E%2B%5D%20%5CRightarrow%20%5BH_3O%5E%2B%5D%3D10%5E%7B-pH%7D%20%5C%5C%20%5C%5C%0ApH%3D2.8%20%5C%5C%0A%5C%20%5BH_3O%5E%2B%5D%3D10%5E%7B-2.8%7D%20%5Capprox%201.58%20%5Ctimes%2010%5E%7B-3%7D)
The [H₃O⁺] of the solution is approximately 1.58 × 10⁻³ M.
Answer:
<em>it's broken down into carbon dioxide and water </em>
Explanation:
