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Howdy Essay

  • Submitted by: mccloudml
  • on November 13, 2012
  • Category: Arts and Music
  • Length: 870 words

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Below is an essay on "Howdy" from Anti Essays, your source for research papers, essays, and term paper examples.

Chapter 10 odd problems solutions
CHEM1405.P02
Dr. Stankus

Dilution of Solutions
1. What volume of (a) 12.0 M HCl is required to make 2.00 L of 1.00 M HCl?
Mconc x Vconc = Mdil x Vdil
(12.0 M) x (Vconc) = (1.00 M) x (2.00 L)
Vconc=0.167 L
(b) 1.04 M Na2CO3 is required to make 0.500 L of 1.00 M Na2CO3?
Mconc x Vconc = Mdil x Vdil
(1.04 M)   x (Vconc) = (1.00 M) x (0.500 L)
Vconc=0.481 L

3. What volume of concentrated (18.0 M) sulfuric acid would be required to make each of the following?
a. 1.25 L of 6.00 M solution
Mconc x Vconc = Mdil x Vdil
(18.0 M)   x (Vconc) = (6.00 M) x (1.25 L)
Vconc=0.417 L

b. 575 mL of 0.100 M solution
Mconc x Vconc = Mdil x Vdil
(18.0 M)   x (Vconc) = (0.100 M) x (0.575 L)
Vconc=0.00319 L

Acid-Base Titrations
5. Calculate the molarity of an HCl solution if 20.0 mL of it requires 33.2 mL of 0.150 M NaOH for neutralization.




7. Calculate the molarity of a Ca(OH)2 solution if 18.5 mL of it requires 28.2 mL of 0.0302 M HCl for neutralization. The products are CaCl2(aq) and H2O.





9.   A 20-mL sample of gastric fluid is neutralized by 25 mL of 0.10 M NaOH. What is the molarity of HCl in the fluid? Assume that all the acidity of the gastric fluid is due to HCl.

 

11.   How many milliliters of 0.100 M H2SO4 are required to react with 10.3 mL of 0.404 M NaHCO3?
H2SO4 (aq) + 2 NaHCO3 (aq)    Na2SO4 (aq) + 2 H2O + 2 CO2(g)
             
13. How many milliliters of 0.0195 M HCl are required to titrate
a.   25.00 mL of 0.0365 M KOH(aq)
 
b.   10.00 mL of 0.0116 M Ca(OH)2(aq),


c.   20.00 mL of 0.0225 M NH3(aq)

pH and pOH
15.   What is the pH of each of the following solutions?
a. 1.0 x 10-2   M HCl
pH = -log[H3O+]= -log(1.0x10-2) = 2.00

b. 1.0 x 10-4   M HNO3
  pH = -log[H3O+]= -log(1.0x10-4) = 4.00

17. What is the pOH of each of the following solutions?
a. 1.0 x 10-2   M NaOH
pOH = -log[OH-]= -log(1.0x10-2) = 2.00

b. 1.0 x 10-3   M KOH
pOH = -log[OH-]=...

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Howdy. Anti Essays. Retrieved December 16, 2018, from the World Wide Web: http://parimatch-stavka7.com/free-essays/Howdy-348721.html


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