Answer:
- Neutral solutions: concentration of hydronium equals the concentration of hydroxide.
- Acid solutions: concentration of hydronium is greater than the concentration of hydroxide.
- Basic solutions concentration of hydronium is lower than the concentration of hydroxide.
Explanation:
Hello,
It is widely known that the pH of water is 7, therefore the pOH of water is also 7 based on:
In such a way, we can compute the concentration of hydronium and hydroxide ions as shown below:
![pH=-log([H^+])\\](https://tex.z-dn.net/?f=pH%3D-log%28%5BH%5E%2B%5D%29%5C%5C)
![[H^+]=10^{-pH}=10^{-7}=1x10^{-7}M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D10%5E%7B-pH%7D%3D10%5E%7B-7%7D%3D1x10%5E%7B-7%7DM)
![pOH=-log([OH^-])](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29)
![[OH^-]=10^{-pOH}=10^{-7}=1x10^{-7}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D10%5E%7B-pOH%7D%3D10%5E%7B-7%7D%3D1x10%5E%7B-7%7DM)
Thus, we notice that the relationship between the concentration of the hydronium is equal for water or neutral solutions. Moreover, if we talk about acid solutions, pH<OH therefore the concentration of hydronium is greater than the concentration of hydroxide. On the other hand if we talk about basic solutions, pH>OH therefore the concentration of hydronium is lower than the concentration of hydroxide.
Best regards.
Answer:
130,000 cubic centimeters
Explanation:
The volume of a box is its length multiplied by its width multiplied by its height (lwh). 423 times 12 times 25 is 126,900. When multiplying, the number of significant figures is the same as the number with the least number of significant figures. The correct number of significant figures in this example is 2, so 126,900 would need to round up to 130,000 in order to have 2 significant figures.
Answer:
0,72 moles of SO₂ remain
Explanation:
The reaction is:
2SO₂ + O₂ → 2SO₃
Where molecular mass of SO₂ is 64,066g/mol and of SO₃ is 80,066g/mol.
86,0g of SO₂ are:
86,0g × (1mol / 64,066g) = <em>1,34 moles of SO₂</em>.
50,0g of SO₃ are:
50,0g × (1mol / 80,066g) = <em>0,62 moles of SO₃</em>.
Now, as 2 moles of SO₂ produce 2 moles of SO₃, the moles of SO₂ that remain after the reaction are the initial moles of SO₂ - moles of SO₃:
1,34 moles - 0,62 moles =
<em>0,72 moles of SO₂ remain</em>
I hope it helps!
