Answer:
Na sodiops los elementos químicos se representan como el átomo, el núcleo, donde se necuentran los protones y neutrones van dentro, los electrones afuera, girando de forma elíptica alrededor del núcleo. ... por ejemplo el oxígeno numero atómico 8, tiene 8 protones y 8 electrones. el número de neutrones es diferente.
Explanation:
THE MOLECULE HAS A C=C AND AN -OH GROUP, SO IT IS CALLED AN ENE/OL, I.E., AN ENOL. ENOLS CAN BE FORMED ONLY FROM CARBONYL COMPOUNDS WHICH HAVE ALPHA HYDROGENS. THEY CAN BE FORMED BY ACID OR BASE CATALYSIS, AND ONCE FORMED ARE HIGHLY REACTIVE TOWARD ELECTROPHILES, LIKE BROMINE.
Answer:
The number of atoms present in one unit of the following compounds is:
a). Potassium Iodide , KI = 2
b).Sodium Sulfide, 
= 3
c). Silicon Dioxide , 
= 3
d). Carbonic Acid ,
= 6
Explanation:
Atomicity : It is defined as the number of atoms that are present in a given molecule/compound.
Atom : The smallest unit of matter is called atom. For e.g O is atom of oxygen but 
is not an atom , it is molecule of oxygen .
molecule has 2 atoms of Oxygen
Similarly Na , K , Fe are atoms but ,
,
,
are molecules
a).Potassium Iodide
KI = 1 atom of K + 1 atom of I
Total atoms = 2
b) Sodium Sulfide
= 2 atoms of Na + 1 atom of S
Total atoms = 3
c) Silicon Dioxide
= 1 atom of Si + 2 atoms of O
Total atoms = 3
d) Carbonic Acid
= 2 atom of H + 1 atom of C + 3 atom of O
= 2+1+3
Total atoms = 6
Final volume is 400 mL
<span>The moles in MgSO4 is 0.00788 </span><span>mL
</span>
The new concentration is 0.197
Answer:
a. 1.78x10⁻³ = Ka
2.75 = pKa
b. It is irrelevant.
Explanation:
a. The neutralization of a weak acid, HA, with a base can help to find Ka of the acid.
Equilibrium is:
HA ⇄ H⁺ + A⁻
And Ka is defined as:
Ka = [H⁺] [A⁻] / [HA]
The HA reacts with the base, XOH, thus:
HA + XOH → H₂O + A⁻ + X⁺
As you require 26.0mL of the base to consume all HA, if you add 13mL, the moles of HA will be the half of the initial moles and, the other half, will be A⁻
That means:
[HA] = [A⁻]
It is possible to obtain pKa from H-H equation (Equation used to find pH of a buffer), thus:
pH = pKa + log₁₀ [A⁻] / [HA]
Replacing:
2.75 = pKa + log₁₀ [A⁻] / [HA]
As [HA] = [A⁻]
2.75 = pKa + log₁₀ 1
<h3>2.75 = pKa</h3>
Knowing pKa = -log Ka
2.75 = -log Ka
10^-2.75 = Ka
<h3>1.78x10⁻³ = Ka</h3>
b. As you can see, the initial concentration of the acid was not necessary. The only thing you must know is that in the half of the titration, [HA] = [A⁻]. Thus, the initial concentration of the acid doesn't affect the initial calculation.
