Answer:
The initial volume of Ne gas is 261mL
Explanation:
This question can be answered using Ideal Gas Equation;
However, the following are the given parameters
Initial Pressure = 654mmHg
Finial Pressure = 345mmHg
Final Volume = 495mL
Required
Initial Volume?
The question says that Temperature is constant;
This implies that, we'll make use of Boyle's law ideal gas equation which states;
![P_1V_1 = P_2V_2](https://tex.z-dn.net/?f=P_1V_1%20%3D%20P_2V_2)
Where
represent the initial pressure
represent the final pressure
represent the initial temperature
represent the final temperature
![P_1 = 654mmHg\\P_2 = 345mmHg\\V_2 = 495mL](https://tex.z-dn.net/?f=P_1%20%3D%20654mmHg%5C%5CP_2%20%3D%20345mmHg%5C%5CV_2%20%3D%20495mL)
Substitute these values in the formula above;
![654 * V_1 = 345 * 495](https://tex.z-dn.net/?f=654%20%2A%20V_1%20%3D%20345%20%2A%20495)
![654V_1 = 170775](https://tex.z-dn.net/?f=654V_1%20%3D%20170775)
Divide both sides by 654
![\frac{654V_1}{654} = \frac{170775}{654}](https://tex.z-dn.net/?f=%5Cfrac%7B654V_1%7D%7B654%7D%20%3D%20%5Cfrac%7B170775%7D%7B654%7D)
![V_1 = \frac{170775}{654}](https://tex.z-dn.net/?f=V_1%20%3D%20%5Cfrac%7B170775%7D%7B654%7D)
![V_1 = 261.123853211](https://tex.z-dn.net/?f=V_1%20%3D%20261.123853211)
(Approximated)
<em>The initial volume of Ne gas is 261mL</em>
So, what's shown here is the ion product of pure water: that is, the product of the concentrations of hydronium and hydroxide ions in pure water at 25 °C. By this relation, if you know the [H₃O⁺], you can calculate the [OH⁻], and vice-versa.
Since [H₃O⁺] × [OH⁻] = 1.0 × 10⁻¹⁴, [OH⁻] = (1.0 × 10⁻¹⁴)/[H₃O⁺].
Substituting the given [H₃O⁺] as 1.25 × 10⁻² M:
[OH⁻] = (1.0 × 10⁻¹⁴)/(1.25 × 10⁻² M) = 8.0 × 10⁻¹³ M.
Wouldn't it be just dubble the amount witch would be 300.0g water
Answer:
more
Explanation:
seems like you did not finish the questions
Atoms of elements is the answer. Hope this helps