Answer:
False
Explanation:
The statement ; Regardless of any concentration of ammonium solution the precipitate of unknown halide after 0.1M AgNO3 will remain is FALSE
This is Because the remaining concentration of AgNO3 is dependent on the solubility of Ag⁺
Photosynthesis. Because the plants give off oxygen as a waste product. Carbon dioxide then moves from the air into the leaves of plants through tiny openings in the plants leaves.
Answer:
![[H^{+}] = 0.761 \frac{mol}{L}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%200.761%20%5Cfrac%7Bmol%7D%7BL%7D)
![[OH^{-}]=1.33X10^{-14}\frac{mol}{L}](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D1.33X10%5E%7B-14%7D%5Cfrac%7Bmol%7D%7BL%7D)

Explanation:
HCl and HNO₃ both dissociate completely in water. A simple method is to determine the number of moles of proton from both these acids and dividing it by the total volume of solution.
. V_{HCl}(L) \\ n_{H^{+} } from HNO_{3} = [HNO_{3}](\frac{mol}{L}). V_{HNO_{3}}(L)](https://tex.z-dn.net/?f=n_%7BH%5E%7B%2B%7D%20%7D%20from%20HCl%20%3D%20%5BHCl%5D%28%5Cfrac%7Bmol%7D%7BL%7D%29.%20V_%7BHCl%7D%28L%29%20%20%5C%5C%20n_%7BH%5E%7B%2B%7D%20%7D%20from%20HNO_%7B3%7D%20%20%3D%20%5BHNO_%7B3%7D%5D%28%5Cfrac%7Bmol%7D%7BL%7D%29.%20V_%7BHNO_%7B3%7D%7D%28L%29)
Here, n is the number of moles and V is the volume. From the given data moles can be calculated as follows






For molar concentration of hydrogen ions:
![[H^{+}] = \frac{n_{H^{+}}(mol)}{V(L)}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%20%3D%20%5Cfrac%7Bn_%7BH%5E%7B%2B%7D%7D%28mol%29%7D%7BV%28L%29%7D)
![[H^{+}] = \frac{0.761}{1.00}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20%5Cfrac%7B0.761%7D%7B1.00%7D)
![[H^{+}] = 0.761 \frac{mol}{L}](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%200.761%20%5Cfrac%7Bmol%7D%7BL%7D)
From dissociation of water (Kw = 1.01 X 10⁻¹⁴ at 25°C) [OH⁻] can be determined as follows
![K_{w} = [H^{+} ][OH^{-} ]](https://tex.z-dn.net/?f=K_%7Bw%7D%20%3D%20%5BH%5E%7B%2B%7D%20%5D%5BOH%5E%7B-%7D%20%5D)
![[OH^{-}]=\frac{Kw}{[H^{+}] }](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D%5Cfrac%7BKw%7D%7B%5BH%5E%7B%2B%7D%5D%20%7D)
![[OH^{-}]=\frac{1.01X10-^{-14}}{0.761 }](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D%5Cfrac%7B1.01X10-%5E%7B-14%7D%7D%7B0.761%20%7D)
![[OH^{-}]=1.33X10^{-14}\frac{mol}{L}](https://tex.z-dn.net/?f=%5BOH%5E%7B-%7D%5D%3D1.33X10%5E%7B-14%7D%5Cfrac%7Bmol%7D%7BL%7D)
The pH of the solution can be measured by the following formula:
![pH = -log[H^{+} ]](https://tex.z-dn.net/?f=pH%20%3D%20-log%5BH%5E%7B%2B%7D%20%5D)


Answer:
c^21, t^729, CH^56>748, do this to calculate the moceluar mass