<h2>
Answer:
atoms</h2>
Explanation:
The given formula of the compound is 
The formula says
Every mole of contains 
moles of atoms of hydrogen.
Given that number of moles of compound is 
So,the number of moles of hydrogen atoms present is 
Since each mole has 
atoms,
moles has 
atoms.
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 molarity of the sodium hydroxide solution is 1.16 M.
Explanation:
Given: 
= 35 mL, 
= 0.5 M
= 15 mL, 
= ?
Formula used to calculate the molarity of the sodium hydroxide solution is as follows.
Substitute the values into above formula as follows.
Thus, we can conclude that the molarity of the sodium hydroxide solution is 1.16 M.
Answer : The value of 
of the reaction is, -369.2 KJ
Explanation :
Formula used :
where,
= internal energy of the reaction = ?
= enthalpy of the reaction = -184.6 KJ/mole = -184600 J/mole
The balanced chemical reaction is,
when the moles of 
are 2 moles then the reaction will be,
From the given balanced chemical reaction we conclude that,
= change in the moles of the reaction = Moles of product - Moles of reactant = 4 - 4 = 0 mole
R = gas constant = 8.314 J/mole.K
T = temperature = 
Now put all the given values in the above formula, we get:
Therefore, the value of of the reaction is, -369.2 KJ
