0,15 moles of NaOH-------in------------1000ml
x moles of NaOH------------in--------100ml
x = 0,015 moles of NaOH
final volume = 150ml
0,015 moles of NaOH---in-------150ml
x moles of NaOH--------------in-----1000ml
x = 0,1 moles of NaOH
answer: 0,1mol/dm³ (molarity)
Explanation:
1) refine the specimen into fine powder 2) place the smallest amount you can see in the capillary tube 3) set the voltage to increase exponentially to 200 below the predicted temperature, then adjust so that the temperature rises to 20 per minute 4) report the temperature at which the liquid first appears and the temperature at which the last crystal disappears.
Answer:
Explanation:
Due to Coulomb´s law electric force can be described by the formula 
, where K is the Coulomb´s constant (
), 
= Charge 1 (Na+ in this case), 
is the charge 2 (Cl-) and r is the distance between both charges.
Work made by a force is W=F.d and total work produced is the change in energy between final and initial state. this is 
.
so we have ![W=W_{f} -W_{i} =(K\frac{q_{(Na+)}q_{(Cl-)}rf}{r_{f} ^{2}})-(K\frac{q_{(Na+)}q_{(Cl-)}ri}{r_{i} ^{2}})=Kq_{(Na+)}q_{(Cl-)[\frac{1}{{r_{f}}} -\frac{1}{{r_{i}}}]](https://tex.z-dn.net/?f=W%3DW_%7Bf%7D%20-W_%7Bi%7D%20%3D%28K%5Cfrac%7Bq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%7Drf%7D%7Br_%7Bf%7D%20%5E%7B2%7D%7D%29-%28K%5Cfrac%7Bq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%7Dri%7D%7Br_%7Bi%7D%20%5E%7B2%7D%7D%29%3DKq_%7B%28Na%2B%29%7Dq_%7B%28Cl-%29%5B%5Cfrac%7B1%7D%7B%7Br_%7Bf%7D%7D%7D%20-%5Cfrac%7B1%7D%7B%7Br_%7Bi%7D%7D%7D%5D)
Given that ri= 1.1nm= 
and rf= infinite distance
![W=(9x10^{9})(1.6x10^{-19})(-1.6x10^{-19})[\frac{1}{\alpha }-\frac{1}{(1.1x10^{-9})}]=2.1x10^{-19}J](https://tex.z-dn.net/?f=W%3D%289x10%5E%7B9%7D%29%281.6x10%5E%7B-19%7D%29%28-1.6x10%5E%7B-19%7D%29%5B%5Cfrac%7B1%7D%7B%5Calpha%20%7D-%5Cfrac%7B1%7D%7B%281.1x10%5E%7B-9%7D%29%7D%5D%3D2.1x10%5E%7B-19%7DJ)
The answer is (4) at the cathode, where reduction occurs. The Na+ gains one electron and become Na(l). So the reaction occurs at cathode and is reduction reaction.
