In order to calculate the new freezing point, we must first find the depression. This is given by:
ΔT = Kf * b * i
where Kf is the cryscopic constant for the solvent, b is the molarity of the solution in moles per kilogram and i is the van't Hoff factor, which tells us how many ions will be released when a substance is dissolved.
Kf = 1.853, b = 1.5, i = 2 (Na⁺ and Cl⁻)
ΔT = 5.5 °C
The new freezing point will be 0 + 5.5
5.5 °C
Answer:
Rise in volume = 72.7 mL
Explanation:
Given data:
Volume of water = 50.0 mL
Mass of brass piece = 193 g
Rise in volume = ?
Solution:
First of all we will calculate the volume of brass.
d = m/v
d = density
m = mass
v = volume
8.5 g/mL = 193 g/ v
v = 193 g/ 8.5 g/mL
v = 22.71 mL
Rise in volume = volume of water + volume of brass
Rise in volume = 50.0 mL + 22.7 mL
Rise in volume = 72.7 mL
Answer:
3.53*
Explanation:
![K_{c} =\frac{[HI]^{2}[Cl_{2}] }{[HCl]^{2}}=\frac{[5.6*10^{-16} ]^{2} [0.0019]}{[0.13]^{2} }=3.53*10^{-32}](https://tex.z-dn.net/?f=K_%7Bc%7D%20%3D%5Cfrac%7B%5BHI%5D%5E%7B2%7D%5BCl_%7B2%7D%5D%20%7D%7B%5BHCl%5D%5E%7B2%7D%7D%3D%5Cfrac%7B%5B5.6%2A10%5E%7B-16%7D%20%5D%5E%7B2%7D%20%5B0.0019%5D%7D%7B%5B0.13%5D%5E%7B2%7D%20%7D%3D3.53%2A10%5E%7B-32%7D)
Kc is the equilibrium constant calculated as the ration of products over reactants with their stoichiometric coefficients as their exponents. So the balanced chemical equation is important. Solids do not form part of the Kc expression, they do not affect the equilibrium constant which is why Iodine is not part of the calculations
Answer:
The mass of (NH₄)₂S we may add to the solution is: 1.98 g
Explanation:
We interpret the given data:
0.114 m → moles of solute in 1kg of solvent
255 g → the mass of solvent
As we have 0.114 moles of solute ((NH₄)₂S) in 1kg, we must determine the moles in our mass of solvent (255 g H₂O)
We convert the mass of solvent to kg → 255 g . 1 kg/1000g = 0.255 kg
Now we can determine the moles of solute, we used:
0.114 mol/kg . 0.255 kg = 0.02907 moles
If we convert the moles to mass → 0.02907 mol . 68.1g /1mol = 1.98 g
