1.2*10^24# atoms of chlorine
Explanation:
Chlorine gas (#Cl_2#) has two atoms of elemental chlorine in a molecule, so:
#1# mol of #Cl_2# have #6*10^23# molecules of #Cl_2#
#1# molecule of #Cl_2# have #2# atoms per molucule
Then #2*6*10^23 = 1.2*10^24# atoms of chlorine in a mol of chlorine gas
The atomic number of an element is based on the number of protons in the atomic nuclei of its atoms.
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:
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