953.6 g of iron (III) oxide (Fe₂O₃)
Explanation:
We have the following chemical reaction:
2 Fe₂O₃ (s) + 3 C (s) → 4 Fe (s) + 3 CO₂ (g)
We calculate the number of moles of CO₂ by using the following formula:
pressure × volume = number of moles × gas constant × temperature
number of moles = (pressure × volume) / (gas constant × temperature)
number of moles of CO₂ = (2.1 × 100) / (0.082 × 300)
number of moles of CO₂ = 8.54 moles
Taking in account the chemical reaction we devise the following reasoning:
if 2 mole of Fe₂O₃ produces 3 mole of CO₂
then X moles of Fe₂O₃ produces 8.54 mole of CO₂
X = (2 × 8.54) / 3 = 5.69 moles of Fe₂O₃
number of moles = mass / molar weight
mass = number of moles × molar weight
mass of Fe₂O₃ = 5.69 × 160 = 953.6 g
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number of moles
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It will result in an increase in the rate of rxn
Answer:
A chemical change occurs when the reactants chemical compositions have changed
Explanation:
a change in colour, change in temperature, change in smell, formation of a precipitate, or the formation of gas bubbles
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Answer:
4.96E-8 moles of Cu(OH)2
Explanation:
Kps es the constant referring to how much a substance can be dissolved in water. Using Kps, it is possible to know the concentration of weak electrolytes. Then, pKps is the minus logarithm of Kps.
Now, we know that sodium hydroxide (NaOH) is a strong electrolyte, who is completely dissolved in water. Therefore the pH depends only on OH concentration originating from NaOH. Let us to figure out how much is that OH concentration.
![pH= -log[H]\\pH= -log (\frac{kw}{[OH]})](https://tex.z-dn.net/?f=pH%3D%20-log%5BH%5D%5C%5CpH%3D%20-log%20%28%5Cfrac%7Bkw%7D%7B%5BOH%5D%7D%29)
![8.23 = - log(\frac{Kw}{[OH]} \\10^{-8.23} = Kw/[OH]\\ [OH] = Kw/10^{-8.23}](https://tex.z-dn.net/?f=8.23%20%3D%20-%20log%28%5Cfrac%7BKw%7D%7B%5BOH%5D%7D%20%5C%5C10%5E%7B-8.23%7D%20%3D%20Kw%2F%5BOH%5D%5C%5C%20%5BOH%5D%20%3D%20Kw%2F10%5E%7B-8.23%7D)
![[OH]=1.69E-6](https://tex.z-dn.net/?f=%5BOH%5D%3D1.69E-6)
This concentration of OH affects the disociation of Cu(OH)2. Let us see the dissociation reaction:

In the equilibrum, exist a concentration of OH already, that we knew, and it will be added that from dissociation, called "s":
The expression for Kps is:
![Kps= [Cu^{2+}] [OH]^2](https://tex.z-dn.net/?f=Kps%3D%20%5BCu%5E%7B2%2B%7D%5D%20%5BOH%5D%5E2)
The moles of (CuOH)2 soluble are limitated for the concentration of OH present, according to the next equation.

"s" is the soluble quantity of Cu(OH)2.
The solution for this third grade equation is 
Now, let us calculate the moles in 1 L:
