Answer:
The molarity of urea in this solution is 6.39 M.
Explanation:
Molarity (M) is <em>the number of moles of solute in 1 L of solution</em>; that is

To calculate the molality, we need to know the number of moles of urea and the volume of solution in liters. We assume 100 grams of solution.
Our first step is to calculate the moles of urea in 100 grams of the solution,
using the molar mass a conversion factor. The total moles of 100g of a 37.2 percent by mass solution is
60.06 g/mol ÷ 37.2 g = 0.619 mol
Now we need to calculate the volume of 100 grams of solution, and we use density as a conversion factor.
1.032 g/mL ÷ 100 g = 96.9 mL
This solution contains 0.619 moles of urea in 96.9 mL of solution. To express it in molarity, we need to calculate the moles present in 1000 mL (1 L) of the solution.
0.619 mol/96.9 mL × 1000 mL= 6.39 M
Therefore, the molarity of the solution is 6.39 M.
Answer:
A. 2 bonds
Explanation:
If you see a charch of ci20 it would easily show the answer.
I suppose it would be forest because in order to have organic matter the soil needs to be rich and fertile,therefore it is forest.
Answer:
Phosphorus
Sulphur
And chlorine are non metals when bonded covalently gain electronic configuration of argon
Like PCl3
SO2 and HCl
Explanation:
The dissociation of formic acid is:

The acid dissociation constant of formic acid,
is:
![k_a = \frac{[HCOO^{-}] [H^{+}]}{HCOOH}](https://tex.z-dn.net/?f=%20k_a%20%3D%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%20%20%5BH%5E%7B%2B%7D%5D%7D%7BHCOOH%7D%20%20%20%20%20)
Rearranging the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7Bk_a%7D%7B%5BH_%2B%5D%7D%20)
pH = 2.75
![pH = -log[H^{+}]](https://tex.z-dn.net/?f=%20pH%20%3D%20-log%5BH%5E%7B%2B%7D%5D%20)
![[H^{+}]= 10^{-2.75} = 1.78 \times 10^{-3}](https://tex.z-dn.net/?f=%20%5BH%5E%7B%2B%7D%5D%3D%2010%5E%7B-2.75%7D%20%3D%201.78%20%5Ctimes%2010%5E%7B-3%7D%20)


Substituting the values in the equation:
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{k_a}{[H_+]}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7Bk_a%7D%7B%5BH_%2B%5D%7D%20)
![\frac{[HCOO^{-}]}{[HCOOH]} = \frac{1.78\times 10^{-4}}{1.78\times 10^{-3}}](https://tex.z-dn.net/?f=%20%5Cfrac%7B%5BHCOO%5E%7B-%7D%5D%7D%7B%5BHCOOH%5D%7D%20%3D%20%5Cfrac%7B1.78%5Ctimes%2010%5E%7B-4%7D%7D%7B1.78%5Ctimes%2010%5E%7B-3%7D%7D%20%20%20)
Hence, the ratio is
.