Molarity = moles of solute / liters of solution
M = 0.5 / 0.05
M = 10.0 mol/L⁻¹
hope this helps!
Impartial is the word that you want to use
Answer:
2H⁺(aq) + Sr(OH)₂(s) ⟶ Sr²⁺(aq) + 2H₂O(ℓ)
Explanation:
You aren't dumb. You just need more time to learn the concepts.
There are three steps you must follow. You must write the:
- Molecular equation
- Ionic equation
- Net ionic equation
1. Molecular equation
2HBr + Sr(OH)₂ ⟶ SrBr₂ + 2H₂O
To predict the states of the substances, we must remember some solubility rules:
- HBr is a strong acid. It dissociates completely in water.
- Most hydroxides are only slightly soluble. Unless the solution is quite dilute, I would write their states in water as "(s)", i.e., a suspension of the solid in water.
- Salts containing Br⁻ are generally soluble.
Acids and bases react to give salts and water.
Thus, the molecular equation is
2HBr(aq) + Sr(OH)₂(s) ⟶ SrBr₂(aq) + 2H₂O(ℓ)
B. Ionic equation
You write all the soluble substances as ions.
2H⁺(aq)+ 2Br⁻(aq) + Sr(OH)₂(s) ⟶ Sr²⁺(aq) + 2Br⁻(aq) + 2H₂O(ℓ)
C. Net ionic equation
To get the net ionic equation, you cancel the ions that appear on each side of the ionic equation.
2H⁺(aq) + <u>2Br⁻(aq)</u> + Sr(OH)₂(s) ⟶ Sr²⁺(aq) + <u>2Br⁻(aq)</u> + 2H₂O(ℓ)
The net ionic equation is
2H⁺(aq) + Sr(OH)₂(s) ⟶ Sr²⁺(aq) + 2H₂O(ℓ)
Answer:
concentration of ![[O_2]](https://tex.z-dn.net/?f=%5BO_2%5D)
= 0.0124 = 12.4 ×10⁻³ M
concentration of ![[CO]](https://tex.z-dn.net/?f=%5BCO%5D)
= 0.0248 = 2.48 ×10⁻² M
concentration of ![[CO_2]](https://tex.z-dn.net/?f=%5BCO_2%5D)
= 0.4442 M
Explanation:
Equation for the reaction:
⇄ 
+ 
Concentration of 
= 
= 0.469
For our ICE Table; we have:
⇄ +
Initial 0.469 0 0
Change - 2x +2x +x
Equilibrium (0.469-2x) 2x x
K = ![\frac{[CO]^2[O]}{[CO_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BCO%5D%5E2%5BO%5D%7D%7B%5BCO_2%5D%5E2%7D)
K = ![\frac{[2x]^2[x]}{[0.469-2x]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B2x%5D%5E2%5Bx%5D%7D%7B%5B0.469-2x%5D%5E2%7D)
Since the value pf K is very small, only little small of reactant goes into product; so (0.469-2x)² = (0.469)²
![x=\sqrt[3]{1.9929*10^{-6}}](https://tex.z-dn.net/?f=x%3D%5Csqrt%5B3%5D%7B1.9929%2A10%5E%7B-6%7D%7D)
x = 0.0124
∴ at equilibrium; concentration of = 0.0124 = 12.4 ×10⁻³ M
concentration of = 2x = 2 ( 0.0124)
= 0.0248
= 2.48 ×10⁻² M
concentration of = 0.469-2x
= 0.469-2(0.0124)
= 0.469 - 0.0248
= 0.4442 M
