Explanation:
because there are 4 Iodines on the left, we'll put. 4 in front of NaI to balance it. This would result in 4 Na on the left, so we'll put a 2 in front of Sodium Sulfate to balance the right side. Now we have 4 Na and I on both side, as well as 2 Sulfate on both sides. Pb is already balanced. The equation is now complete.
1. Berkelium(Berkeley, CA) 2. Dubnium(Dubna, Russia) 3. Darmstaditum (Darmstadt, Germany) 4. Erbium(Ytterby, Sweden) 5. Strontium(Strontian, Scotland) 6. Terbium(Ytterby, Sweden) 7. Yttebium(Ytterby, Sweden) 8. Yttrium(Ytterby, Sweden)
<u>Answer:</u> The solubility of oxygen at 682 torr is 
<u>Explanation:</u>
To calculate the molar solubility, we use the equation given by Henry's law, which is:
Or,
where,
are the initial concentration and partial pressure of oxygen gas
are the final concentration and partial pressure of oxygen gas
We are given:
Conversion factor used: 1 atm = 760 torr
Putting values in above equation, we get:
Hence, the solubility of oxygen gas at 628 torr is
<h3>
Answer:</h3>
0.75 moles NaOH
<h3>
Explanation:</h3>
We are given;
Volume of NaOH solution = 2.5 Liters
Molarity of NaOH = 0.300 M
We are required to calculate the moles of NaOH
We need to establish the relationship between moles, molarity and volume of a solution.
That would be;
Concentration/molarity = Moles ÷ Volume
Therefore;
Moles = Concentration × Volume
Thus;
Moles of NaOH = 0.300 moles × 2.50 L
= 0.75 moles
Therefore, the number of moles of NaOH is 0.75 moles
