Any buffer exists in this equilibrium
HA <=>

In a buffer, there is a large reservoir of both the undissociated acid (HA) and its conjugate base (

)
When a strong acid is added, it reacts with the large reservoir of the conjugate base (

) forming a salt and water. Since this large reservoir of the conjugate base is used, the ph does not alter drastically, but instead resist the pH change.
Copper is an brown-orange color which it's atomic number is 29. With high thermal and electricity conductivity with it's smooth surface.
Answer:
dium (a liquid or a gas). This pattern of motion typically consists of random fluctuations in a particle's position inside a fluid sub-domain, followed by a relocation to another sub-domain. Each relocation is followed by more fluctuations within the new closed volume. This pattern describes a fluid at thermal equilibrium, defined by a given temperature. Within such a fluid, there exists no preferential direction of flow (as in transport phenomena). More specifically, the fluid's overall linear and angular momenta remain null over time. The kinetic energies of the molecular Brownian motions, together with those of molecular rotations and vibrations, sum up to the caloric component of a fluid's internal energy (the Equipartition theorem).
Explanation:
<u>Answer:</u> The solubility of
in water is 
<u>Explanation:</u>
The balanced equilibrium reaction for the ionization of cadmium phosphate follows:

3s 2s
The expression for solubility constant for this reaction will be:
![K_{sp}=[Cd^{2+}]^3[PO_4^{3-}]^2](https://tex.z-dn.net/?f=K_%7Bsp%7D%3D%5BCd%5E%7B2%2B%7D%5D%5E3%5BPO_4%5E%7B3-%7D%5D%5E2)
We are given:

Putting values in above equation, we get:

Hence, the solubility of
in water is 
The answer to your question is full