Answer:
About 0.0940 M.
Explanation:
Recall that NaOH is a strong base, so it dissociates completely into Na⁺ and OH⁻ ions. Because the acid is monoprotic, we can represent it with HA. Thus, the reaction between HA and NaOH is:

Using the fact that it took 15.00 mL of NaOH to reach the endpoint, determine the number of HA that was reacted with:

Therefore, the molarity of the original solution was:
![\displaystyle \left[ \text{HA}\right] = \frac{0.00188\text{ mol}}{20.00\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.0940\text{ M}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cleft%5B%20%5Ctext%7BHA%7D%5Cright%5D%20%3D%20%5Cfrac%7B0.00188%5Ctext%7B%20mol%7D%7D%7B20.00%5Ctext%7B%20mL%7D%7D%20%5Ccdot%20%5Cfrac%7B1000%5Ctext%7B%20mL%7D%7D%7B1%5Ctext%7B%20L%7D%7D%20%3D%200.0940%5Ctext%7B%20M%7D)
In conclusion, the molarity of the unknown acid is about 0.0940 M.
1 liter = 2.11 pints
8 pints = 1 gallon
4900 L * (2.11 pints/1 L) * (1 gallon/8 pints) = 1292.375 gallons
4900 has 2 significant figures, so our final answer should have two. Your final answer is 1300 gallons.
A stock solution is the most concentrated
Umm yo no se no ablo engles
<span>12.4 g
First, calculate the molar masses by looking up the atomic weights of all involved elements.
Atomic weight manganese = 54.938044
Atomic weight oxygen = 15.999
Atomic weight aluminium = 26.981539
Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol
Now determine the number of moles of MnO2 we have
30.0 g / 86.936044 g/mol = 0.345081265 mol
Looking at the balanced equation
3MnO2+4Al→3Mn+2Al2O3
it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So
0.345081265 mol / 3 * 4 = 0.460108353 mol
So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum.
0.460108353 mol * 26.981539 g/mol = 12.41443146 g
Finally, round to 3 significant figures, giving 12.4 g</span>