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.
Answer:
65
Explanation:
estion
Al comenzar la reacción: N2(g) + 2O2(g) ------> 2NO2(g) existe 1 mol de N2 y 2 moles de O2 y al
finalizarla está presente una mezcla formada por 2,2 moles en total, ¿cuál es el rendimiento para la
reacción?
Answer:
P2I4 is otherwise known as "<em>Diphosphorus Tetraiodide</em>"
Answer:
If you put friction in to skating rink then obvioulsy the skaters wont be able to glide and skate.
Explanation:
friction is power that disturbes the moving power. places like skating rink and slides try to have the least amount of friction.
3.9 grams CaCO3
The mass of 2.0 L of water with a density of 1.00 g/ml is 2000 grams.And 1 ppm of that is 2000 / 1000000 = 0.002 grams. So just multiply by the ppm of CaCO3, giving 0.002 g * 1.95x10^3 = 3.90 grams.
Since the least accurate datum we have is 2 significant figures, the result should be rounded to 2 significant figures, giving 3.9 grams.