Balanced equation is
HBr + NaOH ----> NaBr + H2O
Using molar masses
80.912 g HBr reacts with 39.997 g of Naoh to give 18.007 g water
so 1 gram of NaOH reacts with 2.023 g of HBR
and 5.7 reacts with 11.531 g HBr so we have excess HBr in this reaction
Mass of water produced = (5.7 * 18.007 / 39.997 = 2.6 g to 2 sig figs
Answer:
2. All the naturally occurring isotopes of Mg.
Explanation:
You want to know the atomic mass of the magnesium you use in the lab. That’s “natural” magnesium. So, you must use the weighted average of all the naturally occurring isotopes in natural Mg.
1. and 3. are <em>wrong</em>. You won’t get the correct mass for natural Mg if you use only the artificial isotopes for your calculation.
4. is <em>wrong</em>. You must use all the naturally occurring isotopes. The two most abundant isotopes of Mg account for only 90 % of the atoms. If you ignore the other 10 %, your calculation will be wrong.
Answer:
pKa = 3.675
Explanation:
∴ <em>C</em> X-281 = 0.079 M
∴ pH = 2.40
let X-281 a weak acid ( HA ):
∴ HA ↔ H+ + A-
⇒ Ka = [H+] * [A-] / [HA]
mass balance:
⇒<em> C</em> HA = 0.079 M = [HA] + [A-]
⇒ [HA] = 0.079 - [A-]
charge balance:
⇒ [H+] = [A-] + [OH-]... [OH-] is negligible; it comes from to water
⇒ [H+] = [A-]
∴ pH = - log [H+] = 2.40
⇒ [H+] = 3.981 E-3 M
replacing in Ka:
⇒ Ka = [H+]² / ( 0.079 - [H+] )
⇒ Ka = ( 3.981 E-3 )² / ( 0.079 - 3.981 E-3 )
⇒ Ka = 2.113 E-4
⇒ pKa = - Log ( 2.113 E-4 )
⇒ pKa = 3.675
