1 kg/L ----------- 0.001 kg/mL
22.4 kg/L ------- ??
22.4 x 0.001 / 1 => 0.0224 kg/mL
Answer:
if ur gonna cry then just dont cry its simple logic guyss!!!!!!!!!!!!
Explanation:
Answer:
The answer to your question is 25.2 g of acetic acid.
Explanation:
Data
[Acetic acid] = 0.839 M
Volume = 0.5 L
Molecular weight = 60.05 g/mol
Process
1.- Calculate the number of moles of acetic acid
Molarity = moles / volume
-Solve for moles
moles = Molarity x volume
-Substitution
moles = (0.839)(0.5)
-Result
moles = 0.4195
2.- Calculate the mass of acetic acid using proportions and cross multiplications
60.05 g ----------------------- 1 mol
x ----------------------- 0.4195 moles
x = (0.4195 x 60.05) / 1
x = 25.19 g
3.- Conclusion
25.2 g are needed to prepare 0.500 L of Acetic acid 0.839M
Explanation:
Knowing the number of valence electrons in one of the alien elements helps in identifying it because the number of valence electrons can help categorize the alien element. Similar elements have the same valence electrons and knowing the category of the element can help further analyze the element.
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
