In the simulation, open the custom mode. the beaker will be filled to the 0.50 l mark with a neutral solution. set the ph to 11.65 by using the green arrows adjacent to the ph value indicated on the probe in the solution. once you adjust the ph, note the corresponding oh− ion concentration in m as given in the graphic on the left side of the simulation. make sure to select the option "concentration (mol/l)" above the graphic. select on the logarithmic scale below the graphic. find the poh of the solution.

pOH = 2.35

Explanation:

The pH scale is a way to measure the concentration of H⁺ ions in a solution. On the other hand, pOH is a way to refer to the concentration of OH⁻ in a solution. pH and pOH are related through the following equation:

pH + pOH = 14

If pH = 11.65, then

pOH = 14 - pH = 14 -11.65 = 2.35

When pH > 7 and pOH < 7, the solution is basic.

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

the product of the pressure and volume is also constant