Answer:
Kb = 0.428 m/°C
Explanation:
To solve this problem we need to use the <em>boiling-point elevation formula</em>:
- <em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m
Where <em>Tsolution</em> and <em>Tpure solvent</em> are the boiling point of the CS₂ solution (47.52 °C) and of pure CS₂ (46.3 °C), respectively. Kb is the constant asked by the problem, and m is the molality of the solution.
So in order to use that equation and solve for Kb, first we <em>calculate the molality of the solution</em>.
molality = mol solute / kg solvent
- Density of CS₂ = 1.26 g/cm³
- Mass of 410.0 mL of CS₂ ⇒ 410 cm³ * 1.26 g/cm³ = 516.6 g = 0.5166 kg
molality = 0.270 mol / 0.5166 kg = 0.5226 m
Now we <u>solve for Kb</u>:
<em>Tsolution</em> - <em>Tpure solvent</em> = Kb * m
- 47.52 °C - 46.3 °C = Kb * 0.5226 m
The value of X is 10 hence the formula of unknown hydrate sodium sulfate is NaSO4.10 H20
calculation
step 1:find the moles of NaSO4 and the moles of H2O
moles= mass/molar mass
moles of Na2SO4=1.42÷142=0.01 moles
moles of H20= mass of H2O/molar mass of H2O
mass of H2O= 3.22-1.42=1.8g
mole of H2O is therefore 1.8÷18=0.1 moles
step 2: find the mole ratio by dividing each mole by smallest number of mole (0.01)
that is Na2So4= 0.01/0.01 =1
H2O= 0.1/0.01=10
570/8.5=67.0 58... you only have to take the natural part, si the answer is 67 students
Explanation:
It is known that 1 gram contains 1000 milligrams. And, mathematically we can represent it as follows.
or 
So, when we have to convert grams into milligrams then we simply multiply the digit with 1000. And, if we have to convert a digit from milligrams to grams then we simply divide it by 1000.
<u>Answer:</u>
<u>For a:</u> The equilibrium mixture contains primarily reactants.
<u>For b:</u> The equilibrium mixture contains primarily products.
<u>Explanation:</u>
There are 3 conditions:
- When
; the reaction is product favored. - When
; the reaction is reactant favored. - When
; the reaction is in equilibrium.
For the given chemical reactions:
The chemical equation follows:

The expression of
for above reaction follows:
![K_{eq}=\frac{[CN^-][H_3O^+]}{[HCN][H_2O]}=6.2\times 10^{-10}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BCN%5E-%5D%5BH_3O%5E%2B%5D%7D%7B%5BHCN%5D%5BH_2O%5D%7D%3D6.2%5Ctimes%2010%5E%7B-10%7D)
As,
, the reaction will be favored on the reactant side.
Hence, the equilibrium mixture contains primarily reactants.
The chemical equation follows:

The expression of
for above reaction follows:
![K_{eq}=\frac{[HCl]^2}{[H_2][Cl_2]}=2.51\times 10^{4}](https://tex.z-dn.net/?f=K_%7Beq%7D%3D%5Cfrac%7B%5BHCl%5D%5E2%7D%7B%5BH_2%5D%5BCl_2%5D%7D%3D2.51%5Ctimes%2010%5E%7B4%7D)
As,
, the reaction will be favored on the product side.
Hence, the equilibrium mixture contains primarily products.