Answer:
HNO₃ + NaOH ---> NaNO₃ + H₂O
Explanation:
This reaction appears to be a double-displacement reaction. In these reaction, the cation of one compound is swapped with the cation of another. 
As such, the hydrogen cation (H⁺) from HNO₃ is swapped with the sodium cation (Na⁺) of NaOH.
Luckily, all of the cations have a +1 charge and the anions have a -1 charge. This means that no coefficients are necessary to balance the reaction.
The <u>complete balanced </u>equation is:
HNO₃ + NaOH ---> NaNO₃ + H₂O
 
        
             
        
        
        
<span>To work out the volume of something from its density, use the compound measures triangle: mass over density and volume. To find volume that the beaker holds, divide the mass by the density. V = (388.15 - 39.09)/1. V = 349.06g/cm3. To find the weight of the beaker and the contents, first work out the weight (mass) of the mercury, with this formula: mass = d x v. M = 13.5 x 349.06. M = 4712.31. Then add on the weight of the beaker (39.09g). The total weight is 4751.40g.</span>
        
             
        
        
        
42700 milliliters would be the answer...
Hope this helps!
        
             
        
        
        
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
 at 25 °C is  . How to memorize this value?
. 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. 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. 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
 is given. 25 °C implies that  . As a result,
. 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) .
.