Answer:
9.606 grams of citric acid are present in 125 mL of a 0.400 M citric acid solution.
Explanation:
Molarity : It is defined as the number of moles of solute present in one liter of solution. Mathematically written as:
![Molarity=\frac{\text{Moles of solute}}{\text{volume of solution in L}}](https://tex.z-dn.net/?f=Molarity%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20solute%7D%7D%7B%5Ctext%7Bvolume%20of%20solution%20in%20L%7D%7D)
Moles of citric acid = n
Volume of the citric acid solution = 125 mL =125 × 0.001 L= 0.125 L
(1 mL = 0.001L)
Molarity of the citric acid solution = 0.400 M
![0.400 M=\frac{n}{0.12 5L}](https://tex.z-dn.net/?f=0.400%20M%3D%5Cfrac%7Bn%7D%7B0.12%205L%7D)
n = 0.400 M × 0.125 L = 0.05 moles
Mass of 0.05 moles of citric acid :
![0.05 mol\times 192.12 g/mol=9.606 g](https://tex.z-dn.net/?f=0.05%20mol%5Ctimes%20192.12%20g%2Fmol%3D9.606%20g)
9.606 grams of citric acid are present in 125 mL of a 0.400 M citric acid solution.
Answer:
0.04 M
Explanation:
From the question given above, the following data were obtained:
Initial volume of solution (V1) = 25 cm³
Initial Molarity of solution (M1) = 0.4 M
Final volume of solution (V2) = 250 cm³
Final Molarity of solution (M2) =?
We can obtain the final Molarity of the solution by using the dilution formula as illustrated below:
M1V1 = M2V2
0.4 × 25 = M2 × 250
10 = M2 × 250
Divide both side by 250
M2 = 10/250
M2 = 0.04 M
Therefore, the resulting molarity of the diluted solution is 0.04 M
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Hope This Helps! Have A Nice Day!!
Well start by figuring the total amount of buns brought. This is found by multiplying the amount of packages brought by how many there were in each pack...
32 x 12 = 384
Now it says that you had 2 full packs left, and die extra buns. So you can figure how many are left with...
(2 x 12) + 5 = 29
Now subtract the total bins you had by how many you have left, and you'll see how many were used.
384 - 29 = 355
So 355 buns left, tell john he doesn't need so many! I hope this helps!