I believe 212.5m, but I may be wrong, I’m a little rusty with moles
Answer:
It takes 86 days take to cover half of the lake
Explanation:
In the day #1, the amount of the algae is X,
In the day #2 is 2X
In the day #3 is 2*2*X = X*2²
...
In the day #n the amount of the algae is X*2^(n-1)
Assuming X = 1m³. In the day 87, the area infected was:
1m³*2^(87-1)
7.74x10²⁵m³ is the total area of the lake
the half of this amount is 3.87x10²⁵m³
The time transcurred is:
3.87x10²⁵m³ = 1m³*2^(n-1)
Multiplying for 5 in each side:
ln (3.87x10²⁵) = ln (2^(n-1))
58.9175 = n-1 * 0.6931
85 = n-1
86 = n
<h3>It takes 86 days take to cover half of the lake</h3>
Answer:
KOH and H₂SO₄
Explanation:
Neutralization reaction:
It is the reaction in which acid and base react with each other and produce salt and water.
For example:
2KOH + H₂SO₄ → K₂SO₄ + 2H₂O
1. Potassium hydroxide and sulfuric acid react to produce potassium sulfate salt and water.
2. Potassium hydroxide and phosphoric acid react to produce potassium phosphate and water.
H₃PO₄ + 3KOH → K₃PO₄ + 3H₂O
3. Phosphoric acid sodium hydroxide react to produce sodium phosphate and water.
H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
Melting ice cream is a physical change because it is a phase change, the water in the ice cream gains thermal energy and the inter-molecular forces are overcome and the molecules move further apart forming a liquid. It is not a chemical change because no chemical reactions occurred.
Answer: The approximate equilibrium partial pressure of 
is 3.92 atm
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
The given balanced equilibrium reaction is,
![K_p=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
![1.5\times 10^{-5}=\frac{[H_2]^2\times [S_2]}{[H_2S]^2}](https://tex.z-dn.net/?f=1.5%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D%7B%5BH_2S%5D%5E2%7D)
On reversing the reaction:
initial pressure 4.00atm 2.00 atm 0
eqm (4.00-2x)atm (2.00-x) atm 2x atm
![K_p=\frac{[H_2S]^2}{[H_2]^2\times [S_2]}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B%5BH_2S%5D%5E2%7D%7B%5BH_2%5D%5E2%5Ctimes%20%5BS_2%5D%7D)
![0.67\times 10^5=\frac{2x]^2}{[4.00-2x]^2\times [2.00-x]}](https://tex.z-dn.net/?f=0.67%5Ctimes%2010%5E5%3D%5Cfrac%7B2x%5D%5E2%7D%7B%5B4.00-2x%5D%5E2%5Ctimes%20%5B2.00-x%5D%7D)
![[H_2S]=2x=2\times 1.96=3.92 atm](https://tex.z-dn.net/?f=%5BH_2S%5D%3D2x%3D2%5Ctimes%201.96%3D3.92%20atm)
Thus approximate equilibrium partial pressure of is 3.92 atm
