Answer:
It is an area that is covered in grasses and wildflowers that receives enough rainfall to support the grassland but not the growth of trees.
Explanation:
Take a hypothetical sample of exactly 100 grams of the solution.
(16g urea) / (60.06 g urea/mol) = 0.2664 mol urea
((100 g total) - (16g urea)) = 84.0 g H2O = 0.0840 kg H2O
(0.2664 mol) /0.0840 (kg) = 3.17143mol/kg = 3.18m urea
I think it’s tissue I just did it
Answer:
The ratio of acid to conjugate base is outside the buffer range of 10:1.
Explanation:
The Henderson-Hasselbalch equation for a buffer is
![\text{pH} = \text{pK}_{\text{a}} + \log\dfrac{\text{[A$^{-}$]}}{\text{[HA]}}](https://tex.z-dn.net/?f=%5Ctext%7BpH%7D%20%3D%20%5Ctext%7BpK%7D_%7B%5Ctext%7Ba%7D%7D%20%2B%20%5Clog%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%24%5D%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D)
A buffer should have
![\dfrac{1}{10} \leq \dfrac{\text{[A$^{-}]$}}{\text{[HA]}} \leq \dfrac{10}{1}](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B10%7D%20%5Cleq%20%5Cdfrac%7B%5Ctext%7B%5BA%24%5E%7B-%7D%5D%24%7D%7D%7B%5Ctext%7B%5BHA%5D%7D%7D%20%5Cleq%20%5Cdfrac%7B10%7D%7B1%7D)
For a solution that is 1.3 mol·L⁻¹ in HF and 1.3 mmol·L⁻¹ in KF, the ratio is

The ratio of acid to conjugate base is 1000:1, which is outside the range of 10:1.
A is wrong. NF is a weak acid.
C is wrong. The two species are a conjugate acid-base pair.
D is wrong. Salts of Group 1 metals are soluble.
CH₃-CH=CH₂ propene
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