Answer:
<h3>The answer is 7.85 g/mL</h3>
Explanation:
The density of a substance can be found by using the formula

volume = final volume of water - initial volume of water
volume = 13.91 - 12 = 1.91 mL
We have

We have the final answer as
<h3>7.85 g/mL</h3>
Hope this helps you
Answer:
We typically represent covalent bonds with a dash ( - ) between the atoms. This indicates a single bond. Ex: Cl - Cl
Single bond, double bond, triple bond.
Explanation:
We call it a single covalent bond because the atoms are sharing a single pair of electrons.
Answer:
pH = 2.46
Explanation:
Hello there!
In this case, since this neutralization reaction may be assumed to occur in a 1:1 mole ratio between the base and the strong acid, it is possible to write the following moles and volume-concentrations relationship for the equivalence point:

Whereas the moles of the salt are computed as shown below:

So we can divide those moles by the total volume (0.021L+0.0066L=0.0276L) to obtain the concentration of the final salt:
![[salt]=0.01428mol/0.0276L=0.517M](https://tex.z-dn.net/?f=%5Bsalt%5D%3D0.01428mol%2F0.0276L%3D0.517M)
Now, we need to keep in mind that this is an acidic salt since the base is weak and the acid strong, so the determinant ionization is:

Whose equilibrium expression is:
![Ka=\frac{[C_6H_5NH_2][H_3O^+]}{C_6H_5NH_3^+}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BC_6H_5NH_2%5D%5BH_3O%5E%2B%5D%7D%7BC_6H_5NH_3%5E%2B%7D)
Now, since the Kb of C6H5NH2 is 4.3 x 10^-10, its Ka is 2.326x10^-5 (Kw/Kb), we can also write:

Whereas x is:

Which also equals the concentration of hydrogen ions; therefore, the pH at the equivalence point is:

Regards!