Answer:
The balanced equation is 2K(s) + Cl2(g)→2KCl(s)
Fats are large molecules made of two types of molecules, glycerol and some type of fatty acid.
Answer:
* 
* The solution is acidic since the pH is below 7.
Explanation:
Hello,
In this case, we can mathematically define the pH by:
![pH=-log([H_3O^+])](https://tex.z-dn.net/?f=pH%3D-log%28%5BH_3O%5E%2B%5D%29)
Thus, for the given hydronium concentration we simply compute the pH:
Thereby, we conclude the solution is acidic due to the fact that the pH is below 7 which is the neutral point and above it the solutions are basic.
Regards.
<u>Answer:</u>
Those cells that develop differently are referred to Specialised Cells.
<u>
</u><u>Explanation:</u>
Specialised cells are the one that is assigned to perform a specific role. Every specialised cell in the body is assigned to do their own job. The special features in them help them to perform their functions effectively.
Examples of specialised cells are- red blood cells (they are responsible to carry oxygen in the body), nerve cells (specialises in transmitting electrical signals) and muscle cells (brings body parts together).
Answer:
pH =3.8
Explanation:
Lets call the monoprotic weak acid HA, the dissociation equilibria in water will be:
HA + H₂O ⇄ H₃O⁺ + A⁻ with Ka = [ H₃O⁺] x [A⁻]/ [HA]
The pH is the negative log of the H₃O⁺ concentration, we know the equilibrium constant, Ka and the original acid concentration. So we will need to find the [H₃O⁺] to solve this question.
In order to do that lets set up the ICE table helper which accounts for the species at equilibrium:
HA H₃O⁺ A⁻
Initial, M 0.40 0 0
Change , M -x +x +x
Equilibrium, M 0.40 - x x x
Lets express these concentrations in terms of the equilibrium constant:
Ka = x² / (0.40 - x )
Now the equilibrium constant is so small ( very little dissociation of HA ) that is safe to approximate 0.40 - x to 0.40,
7.3 x 10⁻⁶ = x² / 0.40 ⇒ x = √( 7.3 x 10⁻⁶ x 0.40 ) = 1.71 x 10⁻³
[H₃O⁺] = 1.71 x 10⁻³
Indeed 1.71 x 10⁻³ is small compared to 0.40 (0.4 %). To be a good approximation our value should be less or equal to 5 %.
pH = - log ( 1.71 x 10⁻³ ) = 3.8
Note: when the aprroximation is greater than 5 % we will need to solve the resulting quadratic equation.
