Answer:
The value of Ka 
It is a weak acid
Explanation:
From the question we are told that
The concentration of ![[HClO_2]=0.24M](https://tex.z-dn.net/?f=%5BHClO_2%5D%3D0.24M)
The concentration of ![[H^+]=0.051M](https://tex.z-dn.net/?f=%5BH%5E%2B%5D%3D0.051M)
The concentration of ![[ClO_2^-]=0.051M](https://tex.z-dn.net/?f=%5BClO_2%5E-%5D%3D0.051M)
Generally the equation for the ionic dissociation of 
is
The equilibrium constant is mathematically represented as
![= \frac{[H^+][ClO_2^-]}{[HClO_2]}](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7B%5BH%5E%2B%5D%5BClO_2%5E-%5D%7D%7B%5BHClO_2%5D%7D)
Substituting values since all value of concentration are at equilibrium
Since the value of is less than 1 it show that in water it dose not completely
disassociated so it an acid that is weak
Use the formula E=hv, h=plancks constant and v=frequency
use the formula c=v*lambda to find v
the answer will be 2.88*10^-23J
Hydrocarbons
Hope this helps :)
Answer:
4.37 g of barium sulphate
Explanation:
The reaction equation is;
3BaCl2(aq) + Fe2(SO4)3(aq) ---->3 BaSO4(s) + 2FeCl3(aq)
From the question, the number of moles of both barium chloride and FeSO4 = 125/1000 L × 0.150 M = 0.01875 moles
To find the limiting reactant;
3 moles of barium chloride yields 3 moles of barium sulphate
0.01875 moles of barium chloride yields 3 × 0.01875 moles/3 = 0.01875 moles of barium sulphate
1 mole of iron III sulphate yields 3 moles of barium sulphate
0.01875 molesof iron III sulphate yields 0.01875 moles ×3/1 = 0.05625 moles of barium sulphate
Hence,barium chloride is the limiting reactant
Amount of barium sulphate produced = 0.01875 moles × 233 g/mol = 4.37 g of barium sulphate
Answer:
See explanation below
Explanation:
In order to calculate this, we need to use the following expression to get the concentration of the base:
MaVa = MbVb (1)
We already know the volume of NaOH used which is 13.4473 mL. We do not have the concentration of KHP, but we can use the moles. We have the mass of KHP which is 0.5053 g and the molecular formula. Let's calculate the molecular mass of KHP:
Atomic weights of the elements to be used:
K = 39.0983 g/mol; H = 1.0078 g/mol; C = 12.0107 g/mol; O = 15.999 g/mol
MM KHP = (1.0078*5) + (39.0983) + (8*12.0107) + (4*15.999) = 204.2189 g/mol
Now, let's calculate the mole of KHP:
moles = 0.5053 / 204.2189 = 0.00247 moles
With the moles, we also know that:
n = M*V (2)
Replacing in (1):
n = MbVb
Now, solving for Mb:
Mb = n/Vb (3)
Finally, replacing the data:
Mb = 0.00247 / (13.4473/1000)
Mb = 0.184 M
This would be the concentration of NaOH
