Answer:
7
Explanation:
Assume we have 1 L of each solution.
Solution 1
![\text{[H$^{+}$]}= 10^\text{-pH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of H}^{+} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol H}^{+}}{\text{1 L solution}} = 10^{-2}\text{ mol H}^{+}](https://tex.z-dn.net/?f=%5Ctext%7B%5BH%24%5E%7B%2B%7D%24%5D%7D%3D%2010%5E%5Ctext%7B-pH%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%2010%5E%7B%5Ctext%7B-2%7D%7D%20%5Ctext%7B%20mol%2FL%7D%5C%5C%20%5Ctext%7B%20moles%20of%20H%7D%5E%7B%2B%7D%20%3D%20%5Ctext%7B%201%20L%20solution%7D%20%5Ctimes%20%5Cdfrac%7B10%5E%7B-2%7D%5Ctext%7B%20mol%20H%7D%5E%7B%2B%7D%7D%7B%5Ctext%7B1%20L%20solution%7D%7D%20%3D%2010%5E%7B-2%7D%5Ctext%7B%20mol%20H%7D%5E%7B%2B%7D)
Solution 2
pH = 12
pOH = 14.00 - pOH = 14.00 - 12 = 2.0
![\text{[OH$^{-}$]}= 10^\text{-pOH} \text{ mol/L} = 10^{\text{-2}} \text{ mol/L}\\ \text{ moles of OH}^{-} = \text{ 1 L solution} \times \dfrac{10^{-2}\text{ mol OH}^{-}}{\text{1 L solution}} = 10^{-2}\text{ mol OH}^{-}](https://tex.z-dn.net/?f=%5Ctext%7B%5BOH%24%5E%7B-%7D%24%5D%7D%3D%2010%5E%5Ctext%7B-pOH%7D%20%5Ctext%7B%20mol%2FL%7D%20%3D%2010%5E%7B%5Ctext%7B-2%7D%7D%20%5Ctext%7B%20mol%2FL%7D%5C%5C%20%5Ctext%7B%20moles%20of%20OH%7D%5E%7B-%7D%20%3D%20%5Ctext%7B%201%20L%20solution%7D%20%5Ctimes%20%5Cdfrac%7B10%5E%7B-2%7D%5Ctext%7B%20mol%20OH%7D%5E%7B-%7D%7D%7B%5Ctext%7B1%20L%20solution%7D%7D%20%3D%2010%5E%7B-2%7D%5Ctext%7B%20mol%20OH%7D%5E%7B-%7D)
3. pH after mixing
H⁺ + OH⁻ ⟶ H₂O
I/mol: 10⁻² 10⁻²
C/mol: -10⁻² -10⁻²
E/mol: 0 0
The H⁺ and OH⁻ have neutralized each other. The pH will be that of pure water.
pH = 7
Answer:
B.bonds are broken and new bonds are formed
Answer:
1. 2 M
2. 2 M
Explanation:
1. Determination of the final concentration.
Initial Volume (V₁) = 2 L
Initial concentration (C₁) = 6 M
Final volume (V₂) = 6 L
Final concentration (C₂) =?
The final concentration can be obtained as follow:
C₁V₁ = C₂V₂
6 × 2 = C₂ × 6
12 = C₂ × 6
Divide both side by 6
C₂ = 12 / 6
C₂ = 2 M
Therefore, the final concentration of the solution is 2 M
2. Determination of the final concentration.
Initial Volume (V₁) = 0.5 L
Initial concentration (C₁) = 12 M
Final volume (V₂) = 3 L
Final concentration (C₂) =?
The final concentration can be obtained as follow:
C₁V₁ = C₂V₂
12 × 0.5 = C₂ × 3
6 = C₂ × 3
Divide both side by 3
C₂ = 6 / 3
C₂ = 2 M
Therefore, the final concentration of the solution is 2 M
Answer:
Well there is a lot of differences between the two. Its called homogeneous and Heterogeneous mixtures. Homogeneous mixtures are all the substances are evenly distributed throughout the mixture (salt water, air, blood). Heterogeneous mixtures are the substances that are not evenly distributed (chocolate chip cookies, pizza, rocks). So Pasta with sauce and meatballs is heterogeneous and air is homogeneous
HOPE THIS HELPS HAVE A GREAT DAY!!~
Explanation:
