130.954g of Fe2O3 in 0.82 mol
There are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.
HOW TO CALCULATE MASS:
The mass of a substance can be calculated by multiplying the number of moles in the substance by its molecular mass.
However, given that the number of molecules in calcium hydroxide is 4.07 x 10¹⁵molecules, we need to calculate the number of moles in Ca(OH)2 as follows:
no. of moles of Ca(OH)2 = 4.07 x 10¹⁵ ÷ 6.02 × 10²³
no. of moles = 0.676 × 10-⁸
no. of moles = 6.76 × 10-⁹ moles.
Molar mass of Ca(OH)2 = 74.093 g/mol
Mass of Ca(OH)2 = 74.093 × 6.76 × 10-⁹ moles
Mass = 5.01 × 10-⁷grams.
Therefore, there are 5.01 × 10-⁷grams in 4.07 x 10¹⁵molecules of calcium hydroxide.
Learn more about how to calculate mass at: brainly.com/question/8101390?referrer=searchResults
Answer:
Distance of something it can be any type of distance
Answer:
before
1. Prepare a handy emergency supply kit
2. Know the earthquake hazard in your area
3. Familiarize yourself with the exit routes
During
1. Do the DUCK, COVER AND HOLD.
2. stay away from trees and posts.
3. stay clam
4. Don't enter damaged buildings
After
1. Check yourself and others for injuries
2. Monitor news through radio and Tv.
Hope this will help you.
Answer:
d. 0.121 M HC2H3O2 and 0.116 M NaC2H3O2
Explanation:
Hello,
In this case, since the pH variation is analyzed via the Henderson-Hasselbach equation:
![pH=pKa+log(\frac{[Base]}{[Acid]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%20%29)
We can infer that the nearer to 1 the ratio of of the concentration of the base to the concentration of the acid the better the buffering capacity. In such a way, since the sodium acetate is acting as the base and the acetic acid as the acid, we have:
a. ![\frac{[Base]}{[Acid]}=\frac{0.497M}{0.365M}=1.36](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.497M%7D%7B0.365M%7D%3D1.36)
b. ![\frac{[Base]}{[Acid]}=\frac{0.217M}{0.521M}=0.417](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.217M%7D%7B0.521M%7D%3D0.417)
c. ![\frac{[Base]}{[Acid]}=\frac{0.713M}{0.821M}=0.868](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.713M%7D%7B0.821M%7D%3D0.868)
d. ![\frac{[Base]}{[Acid]}=\frac{0.116M}{0.121M}=0.959](https://tex.z-dn.net/?f=%5Cfrac%7B%5BBase%5D%7D%7B%5BAcid%5D%7D%3D%5Cfrac%7B0.116M%7D%7B0.121M%7D%3D0.959)
Therefore, the d. solution has the best buffering capacity.
Regards.
