Chm 151 Week 1
Chapter 11
11.1 Questions
8. Solutions of hydrogen in palladium may be formed by exposing Pd metal to H 2 gas. The H 2 gas applied, but in a concentration of hydrogen in the palladium depends on the pressure of more complex fashion than can be described by Henrys law. Under certain conditions, 0.94g of hydrogen
gas is dissolved in 215 g of palladium metal.
(A) Determine the molarity of this solution (solution density = 1.8
ANSWER: V(Pd) = m/d = 215 g /1.8 3 g/c m cm = 119.4 3 3 g/c m ) = 0.1194 L c = n/V = 0.466 mol/0.1194 L = 3.903 mol/L
(B) Determine the molarity of this solution (solution density = 1.8 g/c m3 ) ANSWER: m= n( H 2 )/3(Pd)=0.466 mol/0.215 kg = 2.17 mol/kg
(C) Determine the percent by mass of hydrogen atoms in this solution (solution density = 1.8
ANSWER: 0.94 g + 215 g = 215.9 g 215.9 g / 1.8 3 g/c m = 119.9 3 cm 3 g/c m ) w/v = (0.94 g / 119.9 c m3 ) x 100% = .0078%
11.2 Questions
12. Compare the process that occur when methanol ( C H 3 OH ¿ , hydrogen chloride (HCl), and the
sodium hydroxide (NaOH) dissolve in water. Write the equations and prepare sketches showing the form
in which each of these compounds is present in its respective solution.
ANSWER: Methanol and the other two will form a homogeneous mixture and will remain as methanol
and water due to no reactivity. Hydrogen chloride will have a decrease of pH due to the acid base −¿
+¿+C l ¿
reaction
The sodium hydroxide will have the opposite, an increase in pH also
H 2 O+ HCl → H 3 O¿
due to the acid base reaction 11.3 Questions −¿
+¿+O H ¿
NaOH ( aq ) → N a¿ 24. The Henry’s law constant for O2 is 1.3 x 10−3 M/atm at 25 ℃ . What mass of oxygen would be dissolved in a 40-L aquarium at 25 ℃ , assuming the atmospheric pressure is 1.00 atm, and
that the partial pressure of O2 is 0.21 atm? 11.4 Questions
30. What are the mole fractions of H 3 PO 4 and water in a solution of 14.5 g of H 3 PO 4 in 125 g of water?
(A) Outline the steps necessary to answer the question
(B) Answer the question
40. The concentration of glucose,
of the solution? C6 H 12 O 6 , in normal spinal fluid is 75 mg
100 g . What is the molarity

• 1493262772-Chm 151 Week 1.docx

SuperTutor

(15)

Status NEW Posted 27 Apr 2017 03:04 AM My Price 12.00

-----------

Attachments

file 1493262783-Solutions file.docx preview (56 words )

S-----------olu-----------tio-----------ns -----------fil-----------e -----------Hel-----------lo -----------Sir-----------/Ma-----------dam----------- T-----------han-----------k y-----------ou -----------for----------- yo-----------ur -----------int-----------ere-----------st -----------and----------- bu-----------yin-----------g m-----------y p-----------ost-----------ed -----------sol-----------uti-----------on.----------- Pl-----------eas-----------e p-----------ing----------- me----------- on----------- ch-----------at -----------I a-----------m o-----------nli-----------ne -----------or -----------inb-----------ox -----------me -----------a m-----------ess-----------age----------- I -----------wil-----------l b-----------e q-----------uic-----------kly----------- on-----------lin-----------e a-----------nd -----------giv-----------e y-----------ou -----------exa-----------ct -----------fil-----------e a-----------nd -----------the----------- sa-----------me -----------fil-----------e i-----------s a-----------lso----------- se-----------nt -----------to -----------you-----------r e-----------mai-----------l t-----------hat----------- is----------- re-----------gis-----------ter-----------ed -----------onÂ----------- th-----------is -----------web-----------sit-----------e. ----------- H-----------YPE-----------RLI-----------NK -----------&qu-----------ot;-----------htt-----------p:/-----------/wo-----------rkb-----------ank-----------247-----------.co-----------m/&-----------quo-----------t; -----------\t -----------&qu-----------ot;-----------_bl-----------ank-----------&qu-----------ot;----------- -----------Tha-----------nk -----------you----------- -----------

Not Rated(0)

• Buy Answer