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:
Colloids are those solution which exhibit Tyndall effect or scatter light.
Answer:
The equilibrium temperature of the coffee is 72.4 °C
Explanation:
Step 1: Data given
Mass of cream = 15.0 grams
Temperature of the cream = 10.0°C
Mass of the coffee = 150.0 grams
Temperature of the coffee = 78.6 °C
C = respective specific heat of the substances( same as water) = 4.184 J/g°C
Step 2: Calculate the equilibrium temperature
m(cream)*C*(T2-T1) = -m(coffee)*c*(T2-T1)
15.0 g* 4.184 J/g°C *(T2 - 10.0°C) = -150.0g *4.184 J/g°C*(T2-78.6°C)
62.76T2 - 627.6 = -627.6T2 + 49329.36
690.36T2 = 49956.96
T2 = 72.4 °C
The equilibrium temperature of the coffee is 72.4 °C
Answer:
Explanation:
We are given the amounts of two reactants, so this is a limiting reactant problem.
1. Assemble all the data in one place, with molar masses above the formulas and other information below them.
Mᵣ: 58.44
NaCl + AgNO₃ ⟶ NaNO₃ + AgCl
m/g: 0.245
V/mL: 50.
c/mmol·mL⁻¹: 0.0180
2. Calculate the moles of each reactant
3. Identify the limiting reactant
Calculate the moles of AgCl we can obtain from each reactant.
From NaCl:
The molar ratio of NaCl to AgCl is 1:1.
From AgNO₃:
The molar ratio of AgNO₃ to AgCl is 1:1.
AgNO₃ is the limiting reactant because it gives the smaller amount of AgCl.
4. Calculate the moles of excess reactant
Ag⁺(aq) + Cl⁻(aq) ⟶ AgCl(s)
I/mmol: 0.900 4.192 0
C/mmol: -0.900 -0.900 +0.900
E/mmol: 0 3.292 0.900
So, we end up with 50. mL of a solution containing 3.292 mmol of Cl⁻.
5. Calculate the concentration of Cl⁻
![\text{[Cl$^{-}$] } = \dfrac{\text{3.292 mmol}}{\text{50. mL}} = \textbf{0.066 mol/L}\\\text{The concentration of chloride ion is $\large \boxed{\textbf{0.066 mol/L}}$}](https://tex.z-dn.net/?f=%5Ctext%7B%5BCl%24%5E%7B-%7D%24%5D%20%7D%20%3D%20%5Cdfrac%7B%5Ctext%7B3.292%20mmol%7D%7D%7B%5Ctext%7B50.%20mL%7D%7D%20%3D%20%5Ctextbf%7B0.066%20mol%2FL%7D%5C%5C%5Ctext%7BThe%20concentration%20of%20chloride%20ion%20is%20%24%5Clarge%20%5Cboxed%7B%5Ctextbf%7B0.066%20mol%2FL%7D%7D%24%7D)
As an object is heated up, its atoms and molecules move and collide faster. Geothermal energy is the thermal energy in the earth. Wind. ... total kinetic energy of the particles hat make up a substance.
