Answer:
B
Explanation:
bonding is a process of two different atoms sharing electrons for stability and these electrons are attracted by one atom losing it's electrons to another
This lesson is the first in a three-part series that addresses a concept that is central to the understanding of the water cycle—that water is able to take many forms but is still water. This series of lessons is designed to prepare students to understand that most substances may exist as solids, liquids, or gases depending on the temperature, pressure, and nature of that substance. This knowledge is critical to understanding that water in our world is constantly cycling as a solid, liquid, or gas.
In these lessons, students will observe, measure, and describe water as it changes state. It is important to note that students at this level "...should become familiar with the freezing of water and melting of ice (with no change in weight), the disappearance of wetness into the air, and the appearance of water on cold surfaces. Evaporation and condensation will mean nothing different from disappearance and appearance, perhaps for several years, until students begin to understand that the evaporated water is still present in the form of invisibly small molecules." (Benchmarks for Science Literacy<span>, </span>pp. 66-67.)
In this lesson, students explore how water can change from a solid to a liquid and then back again.
<span>In </span>Water 2: Disappearing Water, students will focus on the concept that water can go back and forth from one form to another and the amount of water will remain the same.
Water 3: Melting and Freezing<span> allows students to investigate what happens to the amount of different substances as they change from a solid to a liquid or a liquid to a solid.</span>
Answer:
Concentration of OH⁻:
1.0 × 10⁻⁹ M.
Explanation:
The following equilibrium goes on in aqueous solutions:
.
The equilibrium constant for this reaction is called the self-ionization constant of water:
![K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]](https://tex.z-dn.net/?f=K_w%20%3D%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%5Ccdot%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D)
.
Note that water isn't part of this constant.
The value of 
at 25 °C is 
. How to memorize this value?
- The pH of pure water at 25 °C is 7.
![[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-%5Ctext%7BpH%7D%7D%20%3D%2010%5E%7B-7%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
- However,
![[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%3D10%5E%7B-7%7D%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
for pure water. - As a result,
![K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}](https://tex.z-dn.net/?f=K_w%20%3D%20%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%20%5Ccdot%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%2810%5E%7B-7%7D%29%5E%7B2%7D%20%3D%2010%5E%7B-14%7D)
at 25 °C.
Back to this question. ![[\text{H}^{+}]](https://tex.z-dn.net/?f=%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D)
is given. 25 °C implies that 
. As a result,
![\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5B%5Ctext%7BOH%7D%5E%7B-%7D%5D%20%3D%20%5Cfrac%7BK_w%7D%7B%5B%5Ctext%7BH%7D%5E%7B%2B%7D%5D%7D%20%3D%20%5Cfrac%7B10%5E%7B-14%7D%7D%7B1.0%5Ctimes%2010%5E%7B-5%7D%7D%20%3D%2010%5E%7B-9%7D%20%5C%3B%5Ctext%7Bmol%7D%5Ccdot%5Ctext%7Bdm%7D%5E%7B-3%7D)
.
HNO3+KOH = H2O+KNO3 . When nitric acid react with pottasuim hydroxide, the reaction will produce water (H20) and pottasuim trioxonitrate
