Answer:
[H⁺] = 3.16 × 10⁻⁵ mol/L
Explanation:
Given data:
pH of solution = 4.5
Hydrogen ion concentration = ?
Solution;
pH = -log [H⁺]
we will rearrange this formula:
[H⁺] = 10∧-pH
[H⁺] = 10⁻⁴°⁵
[H⁺] = 3.16 × 10⁻⁵ mol/L
Answer:
Explanation:
From the information given:


no of moles of
= 0.01 L × 0.0010 mol/L
no of moles of
= 
no of moles of
= 0.01 L × 0.00010 mol/L
no of moles of
= 
Total volume = 0.02 L
![[Ca^{2+}}] = \dfrac{1\times10^{-5} \ mol}{0.02 \ L} \\ \\ \\ \[[Ca^{2+}}] = 0.0005 \ mol/L](https://tex.z-dn.net/?f=%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%20%5Cdfrac%7B1%5Ctimes10%5E%7B-5%7D%20%5C%20mol%7D%7B0.02%20%5C%20L%7D%20%5C%5C%20%5C%5C%20%20%5C%5C%20%20%5C%5B%5BCa%5E%7B2%2B%7D%7D%5D%20%3D%200.0005%20%5C%20mol%2FL)
![[F^{-}] = \dfrac{(1\times 10^{-6} \ mol)}{0.02 \ L}](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%20%5Cdfrac%7B%281%5Ctimes%2010%5E%7B-6%7D%20%5C%20mol%29%7D%7B0.02%20%5C%20L%7D)
![[F^{-}] = 5 \times 10^{-5} \ mol/L](https://tex.z-dn.net/?f=%5BF%5E%7B-%7D%5D%20%3D%205%20%5Ctimes%2010%5E%7B-5%7D%20%20%5C%20mol%2FL)
![Q = [Ca^{2+}][F^-]^2 \\ \\ Q = 0.0005 \times (5\times 10^{-5})^2 \\ \\ Q = 1.25 \times 10^{-12}](https://tex.z-dn.net/?f=Q%20%3D%20%5BCa%5E%7B2%2B%7D%5D%5BF%5E-%5D%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%200.0005%20%5Ctimes%20%285%5Ctimes%2010%5E%7B-5%7D%29%5E2%20%5C%5C%20%5C%5C%20Q%20%3D%201.25%20%5Ctimes%2010%5E%7B-12%7D)
Since Q<ksp, then there will no be any precipitation of CaF2
Answer:
The answer to your question is V2 = 29.6 l
Explanation:
Data
Pressure 1 = P1 = 12 atm
Volume 1 = V1 = 23 l
Temperature 1 = T1 = 200 °K
Pressure 2 = 14 atm
Volume 2 = V2 = =
Temperature 2 = T2 = 300°K
Process
1.- To solve this problem use the Combine gas law.
P1V1/T1 = P2V2/T2
-Solve for V2
V2 = P1V1T2 / T1P2
2.- Substitution
V2 = (12)(23)(300) / (200)(14)
3.- Simplification
V2 = 82800 / 2800
4.- Result
V2 = 29.6 l
CH3CH2MgBr is more soluble in diethyl ether .
We know that polar solvent dissolve in polar solvent very perfectly . as diethyl ether is a polar solvent so it have dipole -dipole interaction .
Hence the compound with similar interaction can dissolve in diethyl ether .
Here , MgBr2 is an ionic compound . there is ion-ion interactions occurs which is not similar to dipole -dipole interaction in diethyl ether .hence the solubility of MgBr2 in diethyl ether is less .
but in case of CH3CH2MgBr there are both polar and nonpolar end .CH3CH2 is the nonpolar end and MgBr is the polar end .
thus with the nonpolar end solute interact using depression forces and with polar end solute interact using dipole-dipole interaction . so CH3CH2MgBr is more soluble .
Learn more about polar solvent here :
brainly.com/question/3184550
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