Answer:
As with the hydrogen-ion concentration, the concentration of the hydroxide ion can be expressed logarithmically by the pOH. The pOH of a solution is the negative logarithm of the hydroxide-ion concentration. pOH=−log[OH−] The pH of a solution can be related to the pOH.
Answer:
See explanation below
Explanation:
In this case, we have the equilibrium reaction which is:
H₂ + I₂ <------> 2HI Kp = 54
Now, we have the partial pressures of each element in equilibrium, therefore, we can use the expression of equilibrium in this case to calculate the remaining pressure:
Kp = PpHI² / PpH₂ * PpI₂
Solving for the partial pressure of iodine:
PpI₂ = PpHI² / PpH₂ * Kp
Replacing the given values, we have:
PpI₂ = (2.1)² / 0.933 * 54
PpI₂ = 4.41 / 50.382
PpI₂ = 0.088 atm
Answer is: Particles are farther apart in gases so Substance B is a gas.
For example, nitrogen molecules have weakest intermolecular bonds in gas phase and move fast and without order.
Cooling is change from gas to solids. In solid state (for example ice) movement of molecules is more slow than movement of molecules in the gas (for example water vapor).
The process by which a solid changes directly to a gas without first becoming a liquid is called c. sublimation.
Sublimation is an endothermic process. For example dry ice (carbon(IV) oxide in solid state) is used because of sublimation in nightclubs, fog machines, at theaters, haunted house attractions.
Evaporation is when a liquid turns into a gas, while sublimation is when a solid turns into a gas. Both involve absorbing energy to break apart the bonds of the substance. On the other hand, an example of sublimation is dry ice.
Answer: The
of
is
.
Explanation:
Given: ![[Ag^{+}] = 1.3 \times 10^{-4} M](https://tex.z-dn.net/?f=%5BAg%5E%7B%2B%7D%5D%20%3D%201.3%20%5Ctimes%2010%5E%7B-4%7D%20M)
The reaction equation will be written as follows.

This shows that the concentration of
is half the concentration of
ion. So,
![[CrO^{2-}_{4}] = \frac{1.3 \times 10^{-4}}{2}\\= 0.65 \times 10^{-4} M](https://tex.z-dn.net/?f=%5BCrO%5E%7B2-%7D_%7B4%7D%5D%20%3D%20%5Cfrac%7B1.3%20%5Ctimes%2010%5E%7B-4%7D%7D%7B2%7D%5C%5C%3D%200.65%20%5Ctimes%2010%5E%7B-4%7D%20M)
The expression for
of this reaction is as follows.
![K_{sp} = [Ag^{+}]^{2}[CrO^{2-}_{4}]](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5BCrO%5E%7B2-%7D_%7B4%7D%5D)
Substitute values into the above expression as follows.
![K_{sp} = [Ag^{+}]^{2}[CrO^{2-}_{4}]\\= (1.3 \times 10^{-4})^{2} \times 0.65 \times 10^{-4}\\= 1.1 \times 10^{-12}](https://tex.z-dn.net/?f=K_%7Bsp%7D%20%3D%20%5BAg%5E%7B%2B%7D%5D%5E%7B2%7D%5BCrO%5E%7B2-%7D_%7B4%7D%5D%5C%5C%3D%20%281.3%20%5Ctimes%2010%5E%7B-4%7D%29%5E%7B2%7D%20%5Ctimes%200.65%20%5Ctimes%2010%5E%7B-4%7D%5C%5C%3D%201.1%20%5Ctimes%2010%5E%7B-12%7D)
Thus, we can conclude that the
of
is
.