Answer:
a. 63.2%
b. 11.7%
c. 73.3%
d. 0.995%
e. 55.5%
Explanation:
An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).
The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:
*100%
Where 
is the electronegativity difference of the elements. Thus, consulting an electronegativity table:
a. 
= 1.5
= 3.5
*100%
%IC = 63.2%
b. 
= 1.6
= 2.1
*100%
%IC = 11.7%
c. 
= 0.7
= 3.0
*100%
%IC = 73.3%
d. 
= 1.7
= 1.9
*100%
%IC = 0.995 %
e. 
= 1.2
= 3.0
*100%
%IC = 55.5%
Answer:
pH = 4.34
Explanation:
pH= -1/2(logKa) -1/2(log C)
= -1/2( log 5.98*10^-8) -1/2(log 0.0353)
=-1/2(-7.22)-1/2(-1.45)
=3.61+0.725= 4.34
Answer:
Equilibrium constant expression for 
:
![\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{[\mathrm{H_2CO_3}]}{\left[\mathrm{H^{+}\, (aq)}\right]^{2} \, \left[\mathrm{CO_3}^{2-}\right]}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K%20%3D%20%5Cfrac%7B%5Cleft%28a_%7B%5Cmathrm%7BH_2CO_3%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%7B%5Cleft%28a_%7B%5Cmathrm%7BH%5E%7B%2B%7D%7D%7D%5Cright%29%5E2%5C%2C%20%5Cleft%28a_%7B%5Cmathrm%7B%7BCO_3%7D%5E%7B2-%7D%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%20%5Capprox%20%5Cfrac%7B%5B%5Cmathrm%7BH_2CO_3%7D%5D%7D%7B%5Cleft%5B%5Cmathrm%7BH%5E%7B%2B%7D%5C%2C%20%28aq%29%7D%5Cright%5D%5E%7B2%7D%20%5C%2C%20%5Cleft%5B%5Cmathrm%7BCO_3%7D%5E%7B2-%7D%5Cright%5D%7D)
.
Where
, 
, and 
denote the activities of the three species, and ![[\mathrm{H_2CO_3}]](https://tex.z-dn.net/?f=%5B%5Cmathrm%7BH_2CO_3%7D%5D)
, ![\left[\mathrm{H^{+}}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathrm%7BH%5E%7B%2B%7D%7D%5Cright%5D)
, and ![\left[\mathrm{CO_3}^{2-}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cmathrm%7BCO_3%7D%5E%7B2-%7D%5Cright%5D)
denote the concentrations of the three species.
Explanation:
<h3>Equilibrium Constant Expression</h3>
The equilibrium constant expression of a (reversible) reaction takes the form a fraction.
Multiply the activity of each product of this reaction to get the numerator.
is the only product of this reaction. Besides, its coefficient in the balanced reaction is one. Therefore, the numerator would simply be 
.
Similarly, multiply the activity of each reactant of this reaction to obtain the denominator. Note the coefficient "
" on the product side of this reaction. 
is equivalent to 
. The species 
appeared twice among the reactants. Therefore, its activity should also appear twice in the denominator:
.
That's where the exponent "" in this equilibrium constant expression came from.
Combine these two parts to obtain the equilibrium constant expression:
![\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \quad\begin{matrix}\leftarrow \text{from products} \\[0.5em] \leftarrow \text{from reactants}\end{matrix}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K%20%3D%20%5Cfrac%7B%5Cleft%28a_%7B%5Cmathrm%7BH_2CO_3%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%7B%5Cleft%28a_%7B%5Cmathrm%7BH%5E%7B%2B%7D%7D%7D%5Cright%29%5E2%5C%2C%20%5Cleft%28a_%7B%5Cmathrm%7B%7BCO_3%7D%5E%7B2-%7D%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%20%5Cquad%5Cbegin%7Bmatrix%7D%5Cleftarrow%20%5Ctext%7Bfrom%20products%7D%20%5C%5C%5B0.5em%5D%20%5Cleftarrow%20%5Ctext%7Bfrom%20reactants%7D%5Cend%7Bmatrix%7D)
.
<h3 /><h3>Equilibrium Constant of Concentration</h3>
In dilute solutions, the equilibrium constant expression can be approximated with the concentrations of the aqueous "
" species. Note that all the three species here are indeed aqueous. Hence, this equilibrium constant expression can be approximated as:
![\displaystyle K = \frac{\left(a_{\mathrm{H_2CO_3\, (aq)}}\right)}{\left(a_{\mathrm{H^{+}}}\right)^2\, \left(a_{\mathrm{{CO_3}^{2-}\, (aq)}}\right)} \approx \frac{\left[\mathrm{H_2CO_3\, (aq)}\right]}{\left[\mathrm{H^{+}\, (aq)}\right]^2\cdot \left[\mathrm{{CO_3}^{2-}\, (aq)}\right]}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20K%20%3D%20%5Cfrac%7B%5Cleft%28a_%7B%5Cmathrm%7BH_2CO_3%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%7B%5Cleft%28a_%7B%5Cmathrm%7BH%5E%7B%2B%7D%7D%7D%5Cright%29%5E2%5C%2C%20%5Cleft%28a_%7B%5Cmathrm%7B%7BCO_3%7D%5E%7B2-%7D%5C%2C%20%28aq%29%7D%7D%5Cright%29%7D%20%5Capprox%20%5Cfrac%7B%5Cleft%5B%5Cmathrm%7BH_2CO_3%5C%2C%20%28aq%29%7D%5Cright%5D%7D%7B%5Cleft%5B%5Cmathrm%7BH%5E%7B%2B%7D%5C%2C%20%28aq%29%7D%5Cright%5D%5E2%5Ccdot%20%5Cleft%5B%5Cmathrm%7B%7BCO_3%7D%5E%7B2-%7D%5C%2C%20%28aq%29%7D%5Cright%5D%7D)
.
Answer:
The mass of magnesium that has been consumed, was 6.69 g
Explanation:
The reaction is this one:
3Mg (s) + N₂(g) → Mg₃N₂
3 moles of solid magnesium react with 1 mol of nitrogen, to make 1 mol of magnesium nitride.
If 9.27 grams of nitrogen react, we see that ratio is 1:1, so we make 9.27 grams of nitride.
Mass / Molar mass = Moles
9.27 g / 100.9 g/m = 0.092 moles
If we have 0.092 moles of nitride, ratio between Mg is 1:3 so, the rule of three will be:
1 mol of Nitride was produced by 3 moles of Mg (s)
0.092 moles of nitride were produced by, (0.092 .3)/1 = 0.275 moles
Mass og Mg = 24.3 g/m
Molar mass . Moles = Mass
0.275 m . 24.3g/m = 6.69 g
