Answer:
- <u><em>Yes, 200 ml of fluid can be transferred to a 1-quart container.</em></u>
Explanation:
You must compare the two volumes, 200 ml and 1 quart. If 200 ml is less than or equal to 1 quart, then 200 ml of fluid can be transferred to a 1-quart container, else it is not possible.
To compare, the two volumes must be on the same system of units.
Quarts is a measure of volume equivalent to 1/4 of gallon.
One gallon is approximately 3.785 liters.
3.785 liter = 3.785 liter × 1,000 ml/liter
Then, to convert 1 quart to ml use the unit cancellation method:
- (1/4)gallon × 3.785 liter/gallon × 1,000ml / liter = 946.25 ml
Thus, you get that a 1-quart container has volume of 946.25 ml, which allows that 200ml of fluid be transferred to it.
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:
88.1 moles of H₂
Explanation:
Let's make the reaction:
H₂(g) + CO₂(g) → H₂O(g) + CO(g)
Ratio is 1:1
Therefore, 1 mol of hydrogen reacts with 1 mol of carbon dioxide to produce 1 mol of water and 1 mol of carbon monoxide.
In conclusion, 88.1 moles of water must be produced by 88.1 moles of H₂
The reaction can also be written as an equilbrium
H₂(g) + CO₂(g) ⇄ H₂O(g) + CO(g) Kc
