Maybe this example could help you to understand this problem.
https://image.slidesharecdn.com/121howmanyatoms-091201144624-phpapp02/95/12-1-how-many-atoms-17-728....
Like this? 234+34.1= 268.1 then round. If it is less than 5 then you round down if it is more then you round up. Because it is less the final number would be 268.1=268
<h3>
Answer:</h3>
The root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.
<h3>
Solution and Explanation:</h3>
- To find how fast molecules or particles of gases move at a particular temperature, the root mean square speed is calculated.
- Root mean square speed of a gas is calculated by using the formula;

Where R is the molar gas constant, T is the temperature and M is the molar mass of gas in Kg.
<h3>Step 1: Root mean square speed from O₂</h3>
Molar mass of Oxygen is 32.0 g/mol or 0.032 kg/mol
Temperature = 65 degrees Celsius or 338 K
Molar gas constant = 8.3145 J/k.mol


<h3>
Step 2: Root mean square speed of UF₆ </h3>
The molar mass of UF₆ is 352 g/mol or 0.352 kg/mol


Therefore; the root mean square speeds of O₂ and UF₆ is 513m/s and 155 m/s respectively.
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