Answer:
it have 10 neutrons and it is Fluorine(F).
The answer is D
Explanation: A RED FLOWER WITH ONE PART COLORED YELLOW
Answer:
92.75%
Explanation:
The overall chemical equation for the reaction in the preparation of alum from the aluminium can be expressed as:

From above; we will see that 2 moles of Aluminium react with sulphuric acid and water to produce 2 moles o aluminium alum.
Therefore, the theoretical yield can be determined as:

= 4.789g of 
To find the percent yield, we need to divide the actual yield by the theoretical yield and then multiply it with 100.
∴
percent yield = ( mass of alum(g)/theoretical yield(g) ) × 100
percent yield = ( 4.789g / 5.1629g ) × 100%
percent yield = 0.9275 × 100%
percent yield = 92.75%
Thus, the percent yield of the experiment 92.75%
Because Boron has 3 valence electrons it will need to lose valence electrons to have a (in this case) a empty valence shell. This means that it will lose 3 valence electrons to become stable
Answer:
.
.
Explanation:
Hello,
In this case, since the ionization of methylamine is:

The equilibrium expression is:
![Kb=\frac{[CH_3NH_3^+][OH^-]}{[CH_3NH_2]}](https://tex.z-dn.net/?f=Kb%3D%5Cfrac%7B%5BCH_3NH_3%5E%2B%5D%5BOH%5E-%5D%7D%7B%5BCH_3NH_2%5D%7D)
And in terms of the reaction extent
which is equal to the concentration of OH⁻ as well as that of CH₃NH₃⁺ via ice procedure we can write:

Whose solution for
via quadratic equation is 9.24x10⁻³ M since the other solution is negative so it is avoided. Therefore, the concentration of OH⁻ is:
![[OH^-]=x=9.24x10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3Dx%3D9.24x10%5E%7B-3%7DM)
With which we can compute the pOH at first:
![pOH=-log([OH^-])=-log(9.24x10^{-3})=2.034](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%289.24x10%5E%7B-3%7D%29%3D2.034)
Then, since pH and pOH are related via:

The pH turns out:

Best regards.