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
Explanation:
Ions form when atoms gain or lose electrons. This is so that they form a full outer shell of electrons. When an atom gains electrons it becomes a negative ion, because electrons are negatively charged. For example, all halogens (group 7 or 17) form negative ions as they gain an electron forming a 1- charge. When an atom loses electrons it becomes a positive ion, as it is losing some negative charge from the electrons. This would be for example, alkali metals (group 1) which lose an electron to form a positive ion with a 1+ charge, (ALL metals form positive ions).
Answer:
The answer to your question is 25.9 g of KCl
Explanation:
Data
Grams of KCl = ?
Volume = 0.75 l
Molarity = 1 M
Formula
Solve for number of moles
Substitution
Number of moles = 1 x 0.75
Simplification
Number of moles = 0.75 moles
Molecular mass KCl = 39 + 35.5 = 34.5
Use proportions to find the grams of KCl
34.5 g of KCl ---------------- 1 mol
x ---------------- 0.75 moles
x = (0.75 x 34.5) / 1
x = 25.9 g of KCl
% Composition = ?
no atom in 2 g of styrene =?
molar mass of strene =104.15
% composition of c= 12/13.008 =.922*100=92.2
% composition of h =1.008/13.008=0.0774*100= 7.74
no gram atom=mass in kg /molar mass=2/104.15 =0.01920 mol
no of gram atom * avogadro's number = 0.0192*6.02 *10( exponent 23) =1.15584
Answer:
The answer is 18.12KJ is required to vaporise 48.7 g of dichloromethane at its boiling point
Explanation:
To solve the above question we have the given variable as follows
ΔHvap = heat of vaporisation of dichloromethane per mole = 31.6KJ/mole
However since the heat of vaporisation is the heat to vaporise one mole of dichloromethane, then, for 48.7 grams of dichloromethane, we have.
The number of moles of dichloromethane present = 48.7/84.93 = 0.573 moles
Therefore, the amount of heat required to vaporise 48.7 grams of dichloromethane at its boiling point is 31.6KJ/mole×0.573moles =18.12KJ
