It is B. Thank you later please and do good on the test!
This is a straightforward dilution calculation that can be done using the equation
where <em>M</em>₁ and <em>M</em>₂ are the initial and final (or undiluted and diluted) molar concentrations of the solution, respectively, and <em>V</em>₁ and <em>V</em>₂ are the initial and final (or undiluted and diluted) volumes of the solution, respectively.
Here, we have the initial concentration (<em>M</em>₁) and the initial (<em>V</em>₁) and final (<em>V</em>₂) volumes, and we want to find the final concentration (<em>M</em>₂), or the concentration of the solution after dilution. So, we can rearrange our equation to solve for <em>M</em>₂:

Substituting in our values, we get
![\[M_2=\frac{\left ( 50 \text{ mL} \right )\left ( 0.235 \text{ M} \right )}{\left ( 200.0 \text{ mL} \right )}= 0.05875 \text{ M}\].](https://tex.z-dn.net/?f=%5C%5BM_2%3D%5Cfrac%7B%5Cleft%20%28%2050%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%5Cleft%20%28%200.235%20%5Ctext%7B%20M%7D%20%5Cright%20%29%7D%7B%5Cleft%20%28%20200.0%20%5Ctext%7B%20mL%7D%20%5Cright%20%29%7D%3D%200.05875%20%5Ctext%7B%20M%7D%5C%5D.)
So the concentration of the diluted solution is 0.05875 M. You can round that value if necessary according to the appropriate number of sig figs. Note that we don't have to convert our volumes from mL to L since their conversion factors would cancel out anyway; what's important is the ratio of the volumes, which would be the same whether they're presented in milliliters or liters.
it is equal to the potential energy of the products
Answer:
Mass = 5.56 g
Explanation:
Given data:
Mass of Cl₂ = 4.45 g
Mass of NaCl produced = ?
Solution:
Chemical equation:
2Cl₂ + 4NaOH → 3NaCl + NaClO₂ + 2H₂O
Number of moles of Cl₂:
Number of moles = mass/molar mass
Number of moles = 4.45 g/ 71 g/mol
Number of moles = 0.063 mol
Now we will compare the moles of Cl₂ with NaCl.
Cl₂ : NaCl
2 : 3
0.063 : 3/2×0.063 =0.095 mol
Mass of NaCl:
Mass = number of moles × molar mass
Mass = 0.095 mol × 58.5 g/mol
Mass = 5.56 g