The balanced chemical reaction is:
<span>2 I2 + KIO3 + 6 HCl ---------> 5 ICl + KCl + 3 H2O
</span>
We are given the amount of the product to be produced from the reaction. This will be the starting point of our calculations.
28.6 g ICl (1 mol / 162.35 g ICl ) ( 2 mol I2 / 5 mol ICl ) ( 253.81 g I2 / 1 mol I2 ) = 17.88 g I2
C. NaOH ammmonia is also an base but not as strong as NaOH
<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
Answer:
4,572.67
Explanation:
11 divided by 1000= 90.90
44= 11x
x=4
4 x 11= 4000
44-51.300=6.300
6.300 x 90.90=572.67
4000 + 572.67 = the answer
B, E, F have 3 significant figures
