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.
<span>Day and night are not exactly of equal length at the time of the March and September equinoxes. The dates on which day and night are each 12 hours occur a few days before and after the equinoxes.</span>
Answer:
RbI<RbBr<RbCl<RbF
Explanation:
As stated in the question, the latice energy depends on the relative size of the ions. When the action size is constant as in the question, the lattice energy now depends on the relative of the anions. The order of increase in ionic sizes among the halide ions is fluoride<Chloride<Bromide<Iodide. This order of increasing size means that the lattice energy will decrease accordingly as shown in the answer.
Answer:
2NaOH + H2So4 》Na2So4+ 2H2O
Explanation:
SODIUM HYDROXIDE, 2NaOH IS (aq)
FULFURIC ACID, H2So4 IS (aq)
SODIUM SULFATE, Na2So4 IS (aq)
WATER,2H20 IS (l)