Question
experimental value fld percent sulfate in alum
Calculations for Part 2 1504 in KAl(SO4)2 1. Experimental value for percent sulfate in alum BA 0.15799 504) H20 12( 18.02 ):
Results and Data Analysis SON Table 1: Determination of the Water of Hydration in Alum Crystals Mass of crucible + cover + al
0 0
Add a comment Improve this question Transcribed image text
Answer #1

son2 & Bath 96-06269 1 mol sai? in Base 233.686 g/mol alum is converted to basoy for 233.38969 Boso 96-0626 g so Present for

Add a comment
Know the answer?
Add Answer to:
experimental value fld percent sulfate in alum Calculations for Part 2 1504 in KAl(SO4)2 1. Experimental...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • percent error #3 Calculations for Part 2 Ou in KAl(SO4)2 1. Experimental value for percent sulfate...

    percent error #3 Calculations for Part 2 Ou in KAl(SO4)2 1. Experimental value for percent sulfate in alum BA 0.15793504) H20 12 ( 18.0a)= 216.24 KAYS0y), 12H20 MM (504)₂ = 64.327 128.00 2 = 192.32 glmor 2. Theoretical value for percent sulfate in alum are monologa dobar odbud 3. Percent error Results and Data Analysis Table 1: Determination of the Water of Hydration in Alum Crystals Mass of crucible + cover + alum 43.10509 - water Mass of crucible +...

  • Calculate the experimental value for percent sulfate in alum. Calculations for Part 2 1. Experimental value...

    Calculate the experimental value for percent sulfate in alum. Calculations for Part 2 1. Experimental value for percent sulfate in alum Results and Data Analysis Table 1: Determination of the Water of Hydration in Alum Crystals 42.565g Mass of crucible +cover+ alum 41.44292 Mass of crucible + cover Mass of alum Mass of crucible + cover + alum after heating 41.95279 0.50189 Mass of anhydrous alum 58.10 Moles of anhydrous alum til 0.6123 Mass of water of crystallization Moles of...

  • theoretical value for percent sulfate in alum 2. Theoretical value for percent sulfate in alum Results and Data Anal...

    theoretical value for percent sulfate in alum 2. Theoretical value for percent sulfate in alum Results and Data Analysis Table 1: Determination of the water of Hydration in Alum Crystals Mass of crucible + cover + alum 43.10509_-l water Mass of crucible + cover 4.09959 Mass of alum 1.00559_ Mass of crucible + cover + alum after heating 42.65019 Mass of anhydrous alum 0.55069 Moles of anhydrous alum 0.55063 Alun 0.002132 mol nem 3 351 21) Mass of water of...

  • 2. Based on your experimental data cal KAl(SO4)2.nH20. in your experimental data calculated the value of...

    2. Based on your experimental data cal KAl(SO4)2.nH20. in your experimental data calculated the value of n in the chemical formula 3. Calculate the theoretical percentage of SO2- in pure KAl(SO4)2.nH2O using a periodic table for atomic weights and the correct literature value for n. 4. Is your sample alum? Use the results of the three tests to support your answer. Discuss the accuracy of your tests and possible sources of experimental error. Mass of crucible and cover (9) s...

  • Based on your experimental data, calculate the percent water in your sample. Based on your experimental...

    Based on your experimental data, calculate the percent water in your sample. Based on your experimental data calculated the value of n in the chemical formula KAI(SO4)2nH2O. It is often concentra standard Solutions concentr: moisture other har can be m purity ar a simple solution Calculate the theoretical percentage of SO42- in pure KAl(SO4)2 nH20 using a periodic able for atomic weights and the correct literature value for n. DATA TABLE NAME: Sharon Clarke Part I: Melting Temperature Test Results...

  • need help with question 1&2, including work/steps DATA ANALYSIS 1. Based on your experimental data, calculate...

    need help with question 1&2, including work/steps DATA ANALYSIS 1. Based on your experimental data, calculate the percent water in your sample. SHOW ALL WORK 2. Based on your experimental data calculated the value of n in the chemical formula KAI(SO4)2.nH2O. SHOW ALL WORK CULTICOUL Final Melting Temperature/C) Trial 1 Initial Melting Temperature (°C) 90:0°C Initial Melting Temperature (°C) 89.6°C 92. 2°C Final Melting Temperature (°C) 93.2°C Trial 2 Part II: Water of Hydration Test Results Mass of crucible and...

  • 4o UNKNOWN NUMBER Data Trial 2 Trial 1 ২৪.6.209 30.2449 30.3309 30 398 Mass of the crucible and cover &&.4...

    4o UNKNOWN NUMBER Data Trial 2 Trial 1 ২৪.6.209 30.2449 30.3309 30 398 Mass of the crucible and cover &&.4369 Mass of crucible, cover and sample 7989 Mass of crucible, cover and salt after first heating Mass of crucible, cover and salt after second heating &.99 Mass of crucible, cover and salt after third heating Calculations CaSOu CesOy Chemical formula of anhydrous salt (given) 21149 1.6989 9-689 Mass of hydrate (sample) Mass of anhydrous salt 04169 Mass of water liberated...

  • Mass of crucible and cover (g) 31.507 Mass of crucible, cover, and alum before heating (g) Mass o...

    Mass of crucible and cover (g) 31.507 Mass of crucible, cover, and alum before heating (g) Mass of crucible, cover, and alum after 1st heating (g) 1 Mass of crucible, cover, and alum after 2nd heating (g) | Mass of crucible, cover, and alum after 3rd heating (g) Mass of anhydrous alum (g) 3 2, C 2 DATA ANALYSIS Based on your experimental data, calculate the percent water in your sample. 2. Based on your experimental data calculated the value...

  • I synthesized Common Alum in my lab. KAl(SO4)212H2O(s) We then had to find the water content...

    I synthesized Common Alum in my lab. KAl(SO4)212H2O(s) We then had to find the water content based on the initial mass of the Common Alum subtracted from the anhydrous mass after heating. I had 1g of the Common Alum and after heating it, I had 0.57g and I was able to find the percent of water content: 43%. 1g - 0.57 = 0.43g 0.43g / 1g = 0.43 x 100 = 43% Now I have to calculate the percentage error...

  • Alum [KAl(SO4)2·xH2O] is used in food preparation, dye fixation, and water purification. To prepare alum, aluminum...

    Alum [KAl(SO4)2·xH2O] is used in food preparation, dye fixation, and water purification. To prepare alum, aluminum is reacted with potassium hydroxide and the product with sulfuric acid. Upon cooling, alum crystallizes from the solution. (a) A 0.4711−g sample of alum is heated to drive off the waters of hydration, and the resulting KAl(SO4)2 weighs 0.2564 g. Determine the value of x and the complete formula of alum. Value of x: Complete Formula: (b) When 0.7339 g of aluminum is used,...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT