Answer: concentration of solutions are expressed in various ways mol/L, mol/dm3, g/L etc
Explanation:
CH3NH2 + HOH ==> CH3NH3^+ + OH^-Which molecule/ion accepts a proton. That is the base. Which molecule/ion donates a proton. That is the acid.
A stable subatomic particle known by the symbol for "proton"
e elementary charge, p, H+, or 1H+ having a positive electric charge. Its mass is 1,836 times greater than an electron's mass and just a little bit less than that of a neutron (the proton–electron mass ratio). "Nucleons" refers to protons and neutrons together, each of which has a mass of roughly one atomic mass unit (particles present in atomic nuclei).
Each atom. has a nucleus. that contains one or more protons. In order to keep the atomic electrons bound, they offer the central attractive electrostatic force. An element's defining characteristic, known as the atomic number, is the number of protons in the nucleus (represented by the symbol Z)
Learn more about proton here:
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Answer:
if its a neutral atom there should be 6 electrons because there are 6 protons
Explanation:
Answer:
Hornfels most commonly form in the aureole of granitic intrusions in the upper or middle crust. Hornfels formed from contact metamorphism by volcanic activity very close to the surface can produce unusual and distinctive minerals.
<u>Answer:</u> The concentration of
required will be 0.285 M.
<u>Explanation:</u>
To calculate the molarity of
, we use the equation:

Moles of
= 0.016 moles
Volume of solution = 1 L
Putting values in above equation, we get:

For the given chemical equations:

![Ni^{2+}(aq.)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K_f=1.2\times 10^9](https://tex.z-dn.net/?f=Ni%5E%7B2%2B%7D%28aq.%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK_f%3D1.2%5Ctimes%2010%5E9)
Net equation: ![NiC_2O_4(s)+6NH_3(aq.)\rightleftharpoons [Ni(NH_3)_6]^{2+}+C_2O_4^{2-}(aq.);K=?](https://tex.z-dn.net/?f=NiC_2O_4%28s%29%2B6NH_3%28aq.%29%5Crightleftharpoons%20%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%2BC_2O_4%5E%7B2-%7D%28aq.%29%3BK%3D%3F)
To calculate the equilibrium constant, K for above equation, we get:

The expression for equilibrium constant of above equation is:
![K=\frac{[C_2O_4^{2-}][[Ni(NH_3)_6]^{2+}]}{[NiC_2O_4][NH_3]^6}](https://tex.z-dn.net/?f=K%3D%5Cfrac%7B%5BC_2O_4%5E%7B2-%7D%5D%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%7D%7B%5BNiC_2O_4%5D%5BNH_3%5D%5E6%7D)
As,
is a solid, so its activity is taken as 1 and so for 
We are given:
![[[Ni(NH_3)_6]^{2+}]=0.016M](https://tex.z-dn.net/?f=%5B%5BNi%28NH_3%29_6%5D%5E%7B2%2B%7D%5D%3D0.016M)
Putting values in above equations, we get:
![0.48=\frac{0.016}{[NH_3]^6}}](https://tex.z-dn.net/?f=0.48%3D%5Cfrac%7B0.016%7D%7B%5BNH_3%5D%5E6%7D%7D)
![[NH_3]=0.285M](https://tex.z-dn.net/?f=%5BNH_3%5D%3D0.285M)
Hence, the concentration of
required will be 0.285 M.