12. Calculate % difference using equation 2. Show set-up. |trail 1 density - trial 2 density 0.961 - 0.979 x100% = (2) % difference = x 100% average denisty 0.97 % Difference:
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- Hi. I am trying to calculate the percent of water in an uknown hydrate. The given data is: mass of empty crucible +lid = 27.19g mass of crucible + lid and hydrate = 28.58g mass of crucible + lid and sample after heating(anhydrous salt) = 27.68g % water in hydrate formula = mass of water/mass of hydrate * 100X Achieve > Students: Register X Assessment - Chem 101 002X nillanlearning.com/courses/55ace5f9-441e-4942-b91a-6dde768c655e/3/k6bu5e/tools/assessment/items/9f6b56a8-92b9-4b ent Check A LGive Up? O Hint OResources Score: 25 Perform the conversions. atm 1.196 atm = 860.8 mmHg = %3D Pa 1386.9 mmHg : atm %3D 287.3 kPa = %3D/alekscgi/x/lsl.exe/1o_u-IgNslkr7j8P3jH-liGmxpvN4ENzvdYC-70kXyMz36BqJhw3sVP20GT-kvSf4F3al O MEASUREMENT AND MATTER Simplifying unit expressions Rewrite this measurement with a simpler unit, if possible. kg-L's 4.2 kg-s Note: If you can simplify the unit at all, it may be possible to make more than one simplification. Be sure your final
- A hot water bath was made by adding about 250-mL of distilled water to a 400-mL beaker. The beaker was placed on a hot plate and heated until the water was about 70oC. 9.982 g KNO3 was added to a 25-mL graduated cylinder. About 8 mL of distilled water was added to the graduated cylinder and it was placed in the hot water bath. The mixture was stirred continuously with a temperature probe and heated until all the KNO3 dissolved. The solution was clear and colorless and the volume of this solution was 12.5 mL. (This initial solution is called Solution 1.) The graduated cylinder was removed from the hot water bath and, while continuing to stir, Solution 1 was allowed to cool. White crystals of KNO3 were first observed at 64.1oC. Solution 1 Mass of KNO3 9.982 g Volume of Solution 12.5 mL Crystallization temperature 64.1 C Calculate the Ksp value for KNO3 at the crystallization temperature.each mEq NaCI =0.058 g Cost of 100 g KCI =P30.0 Cost of 100 g NaCl =P15.00 5 Price the following prescription using Percentage Mark-Up plus Minimum Fee for the first prescription and the Professional Fee method for the second prescription RI Phenacetin %3D5.00 Aspirin = 4.00 Caffeine citrate =1.00 M. Div. Cap # XV Sig. One ti d Given: Percentage Mark-Up =20% Phenacetin = P152.55 per 1.2 kilo Aspirin = P65.00 per kilo Caffeine citrate =DP25.95per 1.2 kiloD2L Assignments X kJ to calories 1 ANSWER 5590 cal 9.79 x 10¹ lab expt.7.HEIC Removed J > Aktiv Chemistr X RESET 5 1000 0.001 [ENG SUB] lab expt.7.HEIC Removed < X G Which one of t x + =S Update: Submit Show All X
- Mon 4:35 PM nnections | Un X * Student Employment | Car X Illinois State University : H X Mind Tap - Cengage Learn X G If a sample of lithium con index.html?deploymentld%3D55750828934189288909969212&elSBN=9781305657571&nbld%3D2199898&snapshotld%3219989. * DTAP Q Search this Use the References to access important values if needed for this question. If a sample of lithium contains 2.50x10 atoms of lithium-7 and 2.00×10³ atoms of lithium-6, what is the percentage of lithium-6 atoms in the sample? Submit Answer 5 question attempts remaining Next Autosaved at 4:32 PM Backil VHL Central | Dashboard X Consider the following data. Louisiana State University webassign.net web/Student/Assignment-Responses/last?dep=29518359#question308437_16 YouTube U FX - USTVGO.TV = 1.1 Chemistry in Co... AH = 2 NO(g) N₂(g) + O₂(g) 2 NO₂(g) 2 NO(g) + O₂(g) 2 N₂(g) + O₂(g) = 2 N₂O(g) Use Hess's law to calculate AH for the reaction below. KJ X 2022 Fall MATH 1100 for E X y! ml to I - Yahoo Search Res X Psychology 2e ΔΗ = -180.7 kJ ΔΗ = +113.1 kJ ΔΗ = +163.2 kJ Watch Black Clover... VHL Central | Dash... 3 NO(g) N₂O(g) + NO₂(g)How do I obtain the correct number of sig figs and how do I calculate it using the given formula.
- A English II - Pre X 1-4.1.1 Bone x O Emmaree Her A 1ST STOP -C X E Emmaree Her X A Chemistry O L- Unit 5.1 Te x G 2 the relation x A goformative.com/formatives/60466291040cee5e516940cc 6. 8. 9. 10 11 12 13 122 L 7 A gas is placed in a sealed flask at room temperature. The flask is then heated to 50°C. Describe how the pressure of the gas inside the flask will change 2 The four gases above are under the same conditions. In other words, each container has the same pressu volume, and temperature. Which of the following are also true? HINT: More than 1 possible answer chInbc X (531 X Con X I Bala X Ans X Inbc X CHE X 101 Che X С Ч-С х b My x Unk X Sea X E I ma X app.101edu.co/# Bryant's Gmail Cascadia Canvas Lo... T GSBA Scholarship L... HOMEGROWN TRA... Learn Touch Typing... C The Science of Well... Investor360° ® Login ClickUp Reading list >> Question 7 of 40 Submit Which of the following compounds can form intermolecular hydrogen bonds? A) CH4 B) H2Se C) NH3 D) H2 E) All of these compounds can form hydrogen bonds. 10:36 AM e Type here to search 64°F 小 8/26/2021 (8)A sample sodium is being dried. After 1 hour drying, the mass of the sample was 0.337g. After an additional 30 minutes of drying, the mass of the sample is 0.333g. What is the percent change between the first and second meaasurement? if to be considered "dry", two consecutive measurements must have a percent change <_ +- 1%; is the sample dry?
