TLDR: 6.53x10^5 g NH4ClO4
The stoichiometric coefficients (the numbers in front of the reactants and products) show that Aluminum and Ammonium Perchlorate are consumed at the exact same rate throughout the reaction: 3 parts of one to 3 parts of another.
1.5x10^5 grams of Aluminum, considering that the formula weight of Aluminum is 26.98 g/mol, is equal to 5,559.7 moles of Aluminum. This means that 5,559.7 moles of Ammonium Perchlorate are required to run the reaction to completion.
The formula weight of Ammonium Perchlorate is 117.49 grams a mole, and multiplying it by 5,559.7 moles to react to completion means that 6.53x10^5 grams of Ammonium Perchlorate is required for the reaction.
Answer:
The answer to your question is 24.32 g
Explanation:
Data
Atomic weight = ?
HCl volume = 125 ml
Molarity = 0.2
mass of metal = 0.304 g
Balanced chemical equation
M + 2HCl ⇒ MCl₂ + H₂
Process
1.- Calculate the moles of HCl
Molarity = moles / volume (L)
- Solve for moles
moles = Molarity x volume
moles = 0.2 x 0.125
= 0.025
2.- Calculate the moles of the Metal
1 mol of M ----------------- 2 moles of HCl
x ----------------- 0.025 moles of HCl
x = (0.025 x 1) / 2
x = 0.0125 moles of HCl
3.- Calculate the atomic weight of the metal
atomic weight ---------------- 1 mol
0.304 g ---------------0.0125 moles
Atomic weight = (1 x 0.304) / 0.0125
Atomic weight = 24.32 g
Answer: The pH of 0.10 M 
is 4.49.
Explanation:
Given: Initial concentration of 
= 0.10 M
Let us assume that amount of dissociates is x. So, ICE table for dissociation of is as follows.
![Cu(H_{2}O)^{2+}_{6} \rightleftharpoons [Cu(H_{2}O)_{5}(OH)]^{+} + H_{3}O^{+}](https://tex.z-dn.net/?f=Cu%28H_%7B2%7DO%29%5E%7B2%2B%7D_%7B6%7D%20%5Crightleftharpoons%20%5BCu%28H_%7B2%7DO%29_%7B5%7D%28OH%29%5D%5E%7B%2B%7D%20%2B%20H_%7B3%7DO%5E%7B%2B%7D)
Initial: 0.10 M 0 0
Change: -x +x +x
Equilibrium: (0.10 - x) M x x
As the value of 
is very small. So, it is assumed that the compound will dissociate very less. Hence, x << 0.10 M.
And, (0.10 - x) will be approximately equal to 0.10 M.
The expression for value is as follows.
![K_{a} = \frac{[Cu(H_{2}O)^{2+}_{6}][H_{3}O^{+}]}{[Cu(H_{2}O)^{2+}_{6}]}\\1.0 \times 10^{-8} = \frac{x \times x}{0.10}\\x = 3.2 \times 10^{-5}](https://tex.z-dn.net/?f=K_%7Ba%7D%20%3D%20%5Cfrac%7B%5BCu%28H_%7B2%7DO%29%5E%7B2%2B%7D_%7B6%7D%5D%5BH_%7B3%7DO%5E%7B%2B%7D%5D%7D%7B%5BCu%28H_%7B2%7DO%29%5E%7B2%2B%7D_%7B6%7D%5D%7D%5C%5C1.0%20%5Ctimes%2010%5E%7B-8%7D%20%3D%20%5Cfrac%7Bx%20%5Ctimes%20x%7D%7B0.10%7D%5C%5Cx%20%3D%203.2%20%5Ctimes%2010%5E%7B-5%7D)
Hence, ![[H_{3}O^{+}] = 3.2 \times 10^{-5}](https://tex.z-dn.net/?f=%5BH_%7B3%7DO%5E%7B%2B%7D%5D%20%3D%203.2%20%5Ctimes%2010%5E%7B-5%7D)
Formula to calculate pH is as follows.
![pH = -log [H^{+}]](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D)
Substitute the values into above formula as follows.
![pH = -log [H^{+}]\\= - log (3.2 \times 10^{-5})\\= 4.49](https://tex.z-dn.net/?f=pH%20%3D%20-log%20%5BH%5E%7B%2B%7D%5D%5C%5C%3D%20-%20log%20%283.2%20%5Ctimes%2010%5E%7B-5%7D%29%5C%5C%3D%204.49)
Thus, we can conclude that the pH of 0.10 M is 4.49.
Answer:
photosynthesis
Explanation:
its the first step of the cycle
