<em>Answer:</em>
The equlibrium concentration sof Ca+2 ion willl be 4.9×10∧-3 M
<em>Data Given:</em>
Ksp of CaSO4 = 2.4 × 10∧-5
CaSO4 ⇔ Ca+2 + SO4∧-2
<em>Solution:</em>
Ksp = [Ca+2].[ SO4∧-2]
2.4 × 10∧-5 = [x].[x]= x²
x = 4.9×10∧-3 M
<em>Result:</em>
- The conc. of Ca+2 ion is 4.9×10∧-3 M
Answer:
Explanation:
Since water has a chemical formula of H2O , there will be 2 moles of hydrogen in every mole of water. In one mole of water, there will exist approximately 6.02⋅1023 water molecules.
Answer:
RbOH → Rb⁺ + OH⁻
As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base
Explanation:
Arrhenius theory states that a compound is considered a base, if the compound can generate OH⁻ ions in aqueous solution.
Our compound is the RbOH.
When it is put in water, i can dissociate like this:
RbOH → Rb⁺ + OH⁻
As the hydroxide can gives the OH⁻ in water, it is considered as an Arrhenius's base
<u>Answer:</u> The value of
for the given reaction is 1.435
<u>Explanation:</u>
To calculate the molarity of solution, we use the equation:

Given mass of
= 9.2 g
Molar mass of
= 92 g/mol
Volume of solution = 0.50 L
Putting values in above equation, we get:

For the given chemical equation:

<u>Initial:</u> 0.20
<u>At eqllm:</u> 0.20-x 2x
We are given:
Equilibrium concentration of
= 0.057
Evaluating the value of 'x'

The expression of
for above equation follows:
![K_c=\frac{[NO_2]^2}{[N_2O_4]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BNO_2%5D%5E2%7D%7B%5BN_2O_4%5D%7D)
![[NO_2]_{eq}=2x=(2\times 0.143)=0.286M](https://tex.z-dn.net/?f=%5BNO_2%5D_%7Beq%7D%3D2x%3D%282%5Ctimes%200.143%29%3D0.286M)
![[N_2O_4]_{eq}=0.057M](https://tex.z-dn.net/?f=%5BN_2O_4%5D_%7Beq%7D%3D0.057M)
Putting values in above expression, we get:

Hence, the value of
for the given reaction is 1.435