Answer:
you can simply answer vl\t1=v2/t2
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.
Answer:
The final temperature of the water is 28.98 degree Celsius.
Explanation:
It is given that,
Mass of sample of water, m = 52 grams
Initial temperature, 
Heat absorbed, 
The specific heat of water is 
We need to find the final temperature of the water. The heat absorbed is given by the formula as follows :

So, the final temperature of the water is 28.98 degree Celsius.
Answer:
love triangle?- orr they can figure something out