Answer:
3.55atm
Explanation:
We will apply Boyle's law formula in solving this problem.
P1V1 = P2V2
And with values given in the question
P1=initial pressure of gas = 1.75atm
V1=initial volume of gas =7.5L
P2=final pressure of gas inside new piston in atm
V2=final volume of gas = 3.7L
We need to find the final pressure
From the equation, P1V1 = P2V2,
We make P2 subject
P2 = (P1V1) / V2
P2 = (1.75×7.5)/3.7
P2=3.55atm
Therefore, the new pressure inside the piston is 3.55atm
<h2>
Answer:</h2>
A). Increasing the positive charge of the positively charged object and increasing the negative charge of the negatively charged object.
<h2>
Explanation:</h2>
Answer:
Concentration of OH⁻:
1.0 × 10⁻⁹ M.
Explanation:
The following equilibrium goes on in aqueous solutions:
.
The equilibrium constant for this reaction is called the self-ionization constant of water:
![K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]](https://tex.z-dn.net/?f=K_w%20%3D%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%5Ccdot%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D)
.
Note that water isn't part of this constant.
The value of 
at 25 °C is 
. How to memorize this value?
- The pH of pure water at 25 °C is 7.
![[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-%5Ctext%7BpH%7D%7D%20%3D%2010%5E%7B-7%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
- However,
![[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%3D10%5E%7B-7%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
for pure water. - As a result,
![K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}](https://tex.z-dn.net/?f=K_w%20%3D%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%5Ccdot%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%2810%5E%7B-7%7D%29%5E%7B2%7D%20%3D%2010%5E%7B-14%7D)
at 25 °C.
Back to this question. ![[\text{H}^{+}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D)
is given. 25 °C implies that 
. As a result,
![\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cfrac%7BK_w%7D%7B%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%7D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B1.0%5Ctimes%2010%5E%7B-5%7D%7D%20%3D%2010%5E%7B-9%7D%20%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
