The mass of copper that will be produced from copper oxide is determined as 3.19 g.
<h3>
Mass of copper produced from copper oxide</h3>
The mass of copper that will be produced from copper oxide is calculated as follows;
2Cu + O₂ -----> 2CuO
2 1 2
2(63.5) ------------ 2(79.5)
x ---------------------- 4 g
x = (4 x 2 x 63.5)/(2 x 79.5)
x = 3.19 g
Thus, the mass of copper that will be produced from copper oxide is determined as 3.19 g.
Learn more about copper here: brainly.com/question/24856041
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<span>To determine if changing the egg size affects the drop height. hope this helps</span>
Answer:
0.3023 M
Explanation:
Let Picric acid = 
So, + 
⇄ 
+ 
The ICE table can be given as:
+ ⇄ +
Initial: 0.52 0 0
Change: - x + x + x
Equilibrium: 0.52 - x + x + x
Given that;
acid dissociation constant (
) = 0.42
![K_a = \frac{[H_3O^+][Picric^-]}{H_{picric}}](https://tex.z-dn.net/?f=K_a%20%3D%20%5Cfrac%7B%5BH_3O%5E%2B%5D%5BPicric%5E-%5D%7D%7BH_%7Bpicric%7D%7D)
![0.42 = \frac{[x][x]}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5Bx%5D%7D%7B0.52-x%7D%7D)
![0.42 = \frac{[x]^2}{0.52-x}}](https://tex.z-dn.net/?f=0.42%20%3D%20%5Cfrac%7B%5Bx%5D%5E2%7D%7B0.52-x%7D%7D)
0.42(0.52-x) = x²
0.2184 - 0.42x = x²
x² + 0.42x - 0.2184 = 0 -------------------- (quadratic equation)
Using the quadratic formula;
; ( where +/- represent ± )
= 
= 
= 
OR 
= 
OR 
= 
OR 
= 0.30225 OR - 0.72225
So, we go by the +ve integer that says:
x = 0.30225
x = [ ] = [ ] = 0.3023 M
∴ the value of [H3O+] for an 0.52 M solution of picric acid = 0.3023 M (to 4 decimal places).
