Answer:
The answer to your question is K = 115.37
Explanation:
Data
Nitrogen = N₂ [N₂] = 0.15
Chlorine = Cl₂ [Cl₂] = 0.25
Nitrogen trichloride = NCl₃ [NCl₃] = 0.52
Balanced chemical equation
N₂ + 3Cl₂ ⇒ 2NCl₃
Reactants Elements Products
2 N 2
6 Cl 6
- Equilibrium constant
K = ![\frac{[NCl_{3}^{2}]}{[N_{2}][Cl_{2}]^{3}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BNCl_%7B3%7D%5E%7B2%7D%5D%7D%7B%5BN_%7B2%7D%5D%5BCl_%7B2%7D%5D%5E%7B3%7D%7D)
-Substitution
K = ![\frac{[0.52]^{2}}{[0.15][0.25]^{3}}](https://tex.z-dn.net/?f=%5Cfrac%7B%5B0.52%5D%5E%7B2%7D%7D%7B%5B0.15%5D%5B0.25%5D%5E%7B3%7D%7D)
- Simplification
K = 
- Result
K = 115.37
Answer:
A. 0.064mol
B. 0.85mol
C. 1500mL
Explanation:
A. Molarity = 0.33M
Volume = 195mL = 195/1000 = 0.195L
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 0.33 x 0.195
Mole = 0.064mol
B. Molarity = 1.7M
Volume = 500mL = 500/1000 = 0.5L
Mole =?
Molarity = mole /Volume
Mole = Molarity x Volume
Mole = 1.7 x 0.5
Mole = 0.85mol
C. C1 = 12M
V1 = 50mL
C2 = 0.4M
V2 =?
Using the dilution formula C1V1 = C2V2, we find the volume of the diluted solution as follows:
C1V1 = C2V2
12 x 50 = 0.4 x V2
Divide both side by 0.4
V2 = (12 x 50) /0.4
V2 = 1500mL
Answer:
3.6124 m/kg
Explanation:
Molality is calculated as moles of solute (mol) divided by kilogram of solvent (kg). Here, we can find these numbers by using the 35.4%, which gives us 35.4 g of H3PO4 and 100 g of solution to work with.
To go from grams to moles for the phosphoric acid, you need to find the molar mass of the compound or element and divide the grams of the compound or element by that molar mass.
Here, the molar mass for phosphoric acid is 97.9952 g/mol. The equation would look like this:
35.4 g x 1 mol / 97.9952 g = 0.3612422 mol
Next, the 100 g of solvent can easily be converted to 0.1 kg of solvent.
To find the molality, divide the moles of solute and kilograms of solution.
0.3612422 mol / 0.1 kg = 3.6124 m/kg
<span>A metal is one which ionizes easily and gives electrons whencompared to other elements.
So it should have a lower ionization energy.
Hence we would expect element 2 to be a metal.</span>
