Answer:
the value of equilibrium constant for the reaction is 8.5 * 10⁷
Explanation:
Ti(s) + 2 Cl₂(g) ⇄ TiCl₄(l)
equilibrium constant Kc = ![\frac{1}{[Cl_2]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BCl_2%5D%5E2%7D)
Given that,
We are given:
Equilibrium amount of titanium = 2.93 g
Equilibrium amount of titanium tetrachloride = 2.02 g
Equilibrium amount of chlorine gas = 1.67 g
We calculate the No of mole = mass / molar mass
mass of chlorine gas = 1.67 g
Molar mass of chlorine gas = 71 g/mol
mole of chlorine = 1.67 / 71
= 7.0L
Concentration of chlorine is = no of mole / volume
= 0.024 / 7
= 3.43 * 10⁻³M
equilibrium constant Kc =
= ![\frac{1}{[3.43 * 10^-^3]^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5B3.43%20%2A%2010%5E-%5E3%5D%5E2%7D)
= 8.5 * 10⁷
Answer:
The pH is equal to 4.41
Explanation:
Since HClO is a weak acid, its dissociation in aqueous medium is:
HClO ⇄ ClO- + H+
start: 0.05 0 0
change -x +x +x
balance 0.05-x x x
As it is a weak acid it dissociates very little, in its ClO- and H + ions, so the change is negative, where x is a degree of dissociation.
the acidity constant when equilibrium is reached is equal to:
![Ka=\frac{[ClO-]*[H+]}{[HClO]}=\frac{x*x}{0.05-x}=3x10^{-8}](https://tex.z-dn.net/?f=Ka%3D%5Cfrac%7B%5BClO-%5D%2A%5BH%2B%5D%7D%7B%5BHClO%5D%7D%3D%5Cfrac%7Bx%2Ax%7D%7B0.05-x%7D%3D3x10%5E%7B-8%7D)
The 0.05-x fraction can be approximated to 0.05, because the ionized fraction (x) is very small, therefore we have:
clearing the x and calculating its value we have:
![x=3.87x10^{-5}=[H+]=[ClO-]](https://tex.z-dn.net/?f=x%3D3.87x10%5E%7B-5%7D%3D%5BH%2B%5D%3D%5BClO-%5D)
the pH can be calculated by:
![pH=-log[H+]=-log[3.87x10^{-5}]=4.41](https://tex.z-dn.net/?f=pH%3D-log%5BH%2B%5D%3D-log%5B3.87x10%5E%7B-5%7D%5D%3D4.41)
Answer:
The answer to your question is 24.325
Explanation:
Data
Magnesium-24 Abundance = 78.70%
Magnesium-25 Abundance = 10.13%
Magnesium-26 Abundance = 11.17%
Process
1.- Convert the abundance to decimals
Magnesium-24 Abundance = 78.70/100 = 0.787
Magnesium-25 Abundance = 10.13/100 = 0.1013
Magnesium-26 Abundance = 11.17/100 = 0.1117
2.- Write an equation
Average atomic mass = (Atomic mass-1 x Abundance 1) + (Atomic mass 2 x
Abundance-2) + (Atomic mass 3 x Abundance 3)
3.- Substitution
Average atomic mass = (24 x 0.787) + (25 x 0.1013) + (26 x 0.1117)
4.- Simplification
Average atomic mass = 18.888 + 2.533 + 2.904
5.- Result
Average atomic mass = 24.325
For very large numbers, it is much more convenient to use scientific notation. To do this, detect first the position of the decimal point. For whole numbers, the decimal point is place implicitly after the very last digit. Then, move this decimal point to the left until you reach to the last digit. In this case, you moved 8 places until you reach 6.4. Because the number is more than 1, the exponent would have a positive sign. Hence, the scientific notation would be 6.4×10⁸.
