Answer: grams=0.048g, ounces=0.0017oz, 0.00011lb
Explanation:
Stoichiometry
48 mg x 1 g
÷ 1000 mg = 0.048 g
48 mg x 1 g x 16 oz
÷ 1000 mg ÷ 453.6 g = 0.0017 oz
48 mg x 1 g x 1 lb
÷ 1000 mg ÷ 453.6 g = 0.00011 lb
Answer : The correct option is, (D) 100 times the original content.
Explanation :
As we are given the pH of the solution change. Now we have to calculate the ratio of the hydronium ion concentration at pH = 5 and pH = 3
As we know that,
![pH=-\log [H_3O^+]](https://tex.z-dn.net/?f=pH%3D-%5Clog%20%5BH_3O%5E%2B%5D)
The hydronium ion concentration at pH = 5.
![5=-\log [H_3O^+]](https://tex.z-dn.net/?f=5%3D-%5Clog%20%5BH_3O%5E%2B%5D)
..............(1)
The hydronium ion concentration at pH = 3.
![3=-\log [H_3O^+]](https://tex.z-dn.net/?f=3%3D-%5Clog%20%5BH_3O%5E%2B%5D)
................(2)
By dividing the equation 1 and 2 we get the ratio of the hydronium ion concentration.
![\frac{[H_3O^+]_{original}}{[H_3O^+]_{final}}=\frac{1\times 10^{-5}}{1\times 10^{-3}}=\frac{1}{100}](https://tex.z-dn.net/?f=%5Cfrac%7B%5BH_3O%5E%2B%5D_%7Boriginal%7D%7D%7B%5BH_3O%5E%2B%5D_%7Bfinal%7D%7D%3D%5Cfrac%7B1%5Ctimes%2010%5E%7B-5%7D%7D%7B1%5Ctimes%2010%5E%7B-3%7D%7D%3D%5Cfrac%7B1%7D%7B100%7D)
![100\times [H_3O^+]_{original}=[H_3O^+]_{final}](https://tex.z-dn.net/?f=100%5Ctimes%20%5BH_3O%5E%2B%5D_%7Boriginal%7D%3D%5BH_3O%5E%2B%5D_%7Bfinal%7D)
From this we conclude that when the pH of a solution changes from a pH of 5 to a pH of 3, the hydronium ion concentration is 100 times the original content.
Hence, the correct option is, (D) 100 times the original content.
Answer:
a) Heterogeneous mixture (b) Homogenous mixture (c) Pure substance (d) Pure substance
Explanation:
Homogenous mixtures contains mixture of substances with similar proportions while Heterogenous mixture contains substances with a varying proportion.
Answer:
Lewis Structure:
Explanation:
CO2 Lewis Structure
So CO2 = 4 + 6(2) = 16. So, total valence electrons are 16. Carbon is the least electronegative that means it stays at the center. So, put the Carbon in the middle and then set the oxygen either side of that.