Since Na has a 1+ charge and O has a -2 charge, by reversing the charges and placing them as subscripts for the other atoms the formula is Na2O1 or simply Na2O.
Answer:
pOH=9.9
Explanation:
pH=-log[H+]= -log[0.0000877]
=4.06
pOH+ pH=14
pOH=14-4.06= 9.91
Answer:
Heat flux = 13.92 W/m2
Rate of heat transfer throug the 3m x 3m sheet = 125.28 W
The thermal resistance of the 3x3m sheet is 0.0958 K/W
Explanation:
The rate of heat transfer through a 3m x 3m sheet of insulation can be calculated as:

The heat flux can be defined as the amount of heat flow by unit of area.
Using the previous calculation, we can estimate the heat flux:

It can also be calculated as:

The thermal resistance can be expressed as

For the 3m x 3m sheet, the thermal resistance is

Answer:
2Cu2^+ + 2I^- ----> 2Cu^+ + I2
Explanation:
The reaction performed in the experiment is;
2 Cu(NO3)2 + 4 KI → 2 CuI (s) + 4 KNO3 + I2
The iodide ions reduces Cu^2+ to Cu^+ which is insoluble in water hence the precipitate. This is so because iodine is a good oxidizing agent seeing that it requires one electron to fill its outermost shell. Potassium on the other hand is a good reducing agent since it easily looses its one electron.
The oxidation - reduction equation is as follows;
2Cu2^+ + 2e ----> 2Cu^+ reduction half equation
2I^- ----> I2 + 2e. Oxidation half equation
Balanced redox reaction equation;
2Cu2^+ + 2I^- ----> 2Cu^+ + I2