To fully understand the problem, we use the ICE table to identify the concentration of the species. We calculate as follows:
Ka = 2.0 x 10^-9 = [H+][OBr-] / [HOBr]
HOBr = 0.50 M
KOBr = 0.30 M = OBr-
<span> HOBr + H2O <-> H+ + OBr- </span>
<span>I 0.50 - 0 0.30 </span>
<span>C -x x x
</span>---------------------------------------------
<span>E(0.50-x) x (0.30+x) </span>
<span>Assuming that the value of x is small as compared to 0.30 and 0.50 </span>
<span>Ka = 2.0 x 10^-9 = x (0.30) / 0.50) </span>
<span>x = 3.33 x 10^-9 = H+</span>
pH = 8.48
Assuming it has no electrical charge, your electron count would be equal to the atomic number.
Temp must be Kelvin
38 C =
<span>
<span>
<span>
311.15
</span>
</span>
</span>
K
Volume at STP = 8.50 liters * (273.15 / 311.15) * (725 / 760) =
<span>
<span>
<span>
7.1182746306
</span>
</span>
</span>
Liters
The formula to use is:
Volume at STP = Present Volume * (273.15 / Present Temp °K) * (Present Pressure (Torr) / 760)
Answer:
<h3>CH3OH (Methanol)</h3>
MM = 12 + 8 + 16 = 36u
Answer:
I believe it is B (if you get it wrong it's my fault)
