ChromeFileEdit View History Bookmarks ProfilesTab Window Help100%TueWPAL 101_ 23 xPurple/Black6 Gloss Purple E XG s550 mustang x• $550 Mustanc x- (2 unread) - d XO ALEKSALEKS - Davic XA www-awu.aleks.com/alekscgi/x/lsl.exe/1o_u-IgNslkr7j8P3jH-IvdWKW_BBZZI6tTytly4Fcfu6zOtOf8oMM9sefhQp9QAYI8idN-02kvSt27lmROEJa_z..D PausedO CHEMICAL REACTIONSDavidSolving moles-to-moles limiting reactant problems3/5Methane (CH) gas and oxygen (0,) gas react to form carbon dioxide (CO,) gas and water(H,0) vapor. Suppose you have 3.0 mol of CH, and 1.0 mol ofO, in a reactor.Calculate the largest amount of CO, that could be produced. Round your answer to the nearest 0.1 mol.O molExplanationCheck© 2022 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center | AccessibilityII

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Science

Chemistry

) mol

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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https://cxp.cengage.com/contentservice/assets/owms01h/references/chemtables/gen_chem/eo.html ^ table image:
Show all steps leading to the final answer po. Here’s a pdf file in accordance with the topic po: https://drive.google.com/file/d/1_FnDtXCrFKSol3RNWIG_9tNQ7IxgxD6t/view?usp=drivesdk
PLEASE GO THROUGH THE CONCEPT AND ANSWER THE QUESTION ASAP. I need it quick please and I will definitely consider giving thumbs up if you help me. Link: https://drive.google.com/file/d/1mN-k-am1ILxrEOzVQ-8kLtrv5vK8HFvw/view?usp=sharing Question
Chemistry (i) For T = 298K ΔG° = -8.314 x 298 x ln (2.83x10-3) = 14.524 KJ/mol (ii) For T = 308.15K ΔG° = -8.314 x 308.15 x ln (9.619x10-3) = 11.891 KJ/mol (iii) For T = 318.15K ΔG° = -8.314 x 318.15 x ln (3.14x10-2) = 9.147 KJ/mol (iv) For T = 328.15K ΔG° = -8.314 x 328.15 x ln (8.82x10-2) = 6.621 KJ/mol (v) For T = 338.15K ΔG° = -8.314 x 338.15 x ln (2.42x10-1) = 3.990 KJ/mol Based on the measured ΔG° values, is this equilibrium spontaneous at room temperature? Which factor, entropy, or enthalpy, has the greater impact on spontaneity in this case? Explain your answers.
There's 1 drink (and you are asked to determine the glucose concentration in the drink in the units of g/100mL. (Why these units? Well, once you have the concentrations in g/100mL you will be able to compare your values with the nutritional values given on the drink bottles’ labels). The sample of the drink was diluted 1/100 (i.e. by a factor of 100). This was an essential step in the method because, without it, the machine used to analyse the glucose concentration (spectrophotometer) would have given an error as the concentration would have been too high for accurate detection. What this means for you is that the dilution factor will need to be taken into consideration in your calculations (remember the aim is to calculate the concentration in the original drink and not in the diluted drink). You measured the concentration of their diluted drink using the spectrophotometer and their results were provided to them in the units mM (millimolar). Glucose Concentration in mM of drink =…
There's 1 drink (and you are asked to determine the glucose concentration in the drink in the units of g/100mL. (Why these units? Well, once you have the concentrations in g/100mL you will be able to compare your values with the nutritional values given on the drink bottles’ labels). The sample of the drink was diluted 1/100 (i.e. by a factor of 100). This was an essential step in the method because, without it, the machine used to analyse the glucose concentration (spectrophotometer) would have given an error as the concentration would have been too high for accurate detection. What this means for you is that the dilution factor will need to be taken into consideration in your calculations (remember the aim is to calculate the concentration in the original drink and not in the diluted drink). You measured the concentration of their diluted drink using the spectrophotometer and their results were provided to them in the units mM (millimolar). Glucose Concentration in mM of drink =…
Aleks data for AgBrO3 is 5.38 x 10^-5
Sample IdentificationCodeConcentration of M%TA=2-log(%T)Q50004.00 x 10-417.90.75Q50013.20 x 10-425.00.6Q50022.40 x 10-435.70.46Q50031.60 x 10-450.20.3Q50048.000 x 10-570.80.15SampleIdentificationCode%TA=2-log(%T)AMQ0210150143.70.359518560.360.000192Q0210150244.10.355561410.360.00018Q0210150343.80.358525890.360.00017Q0210150444.10.355561410.360.00018Q0210150543.80.358525890.360.00017What was their percent error?43%Does Batch 021015 meet legal requirements?No, because it is not between 2.85 * 10(4) and 3.15 * 10(4)Well #DropsBluedye1234567891012345678910Drops 9Distilled water876543210Concne 0.26tration0.52Test Tube #0.781.041.3Solutions3Concentration (M)2.082.32.6Concentration (ppm)1:1 dilution11.82Starting Dilution21.562:1 dilution0A.Zero standard0Was your calibration curve as linear as you expected?B.Did you experience any â€œdriftâ€ of the resistance readings?C.What is the equation of your best-fit line?D.What commercial drink did you analyze?E.Assuming…
Hot mix asphalt (HMA) is commonly used in the construction of roadways in the United States at a rateestimated to be 500,000,000 tons/year. A typical upper layer (there may be as many as 5 layers ofdifferent thicknesses and compositions) of a high-volume highway is composed of crushed stone (alsoknown as crushed aggregate) and asphalt binder and is approximately 1.25 inches thick. The stone,which has a specific gravity of about 2.7, is mixed with asphalt binder (SG = 1.03) to form the HMAwith a composition of approximately 95 wt% aggregate and 5% asphalt. What is the volume ft3 of asphalt binder and weight (tons) of HMA required for the top layer of a 15-mile segment of interstatehighway with 5 lanes, each of which has a width of 12 ft?
Please answer super fast and answer all questions and show calculations For the image attached For 1. a Mass of metal: Trial 1 is 35.0228 g Trial 2 is 35.0915 g Trial 3 is 34.0821 g Mass of water: Trial 1 is 20.0177 g Trial 2 is 20.0250 g Trial 3 is 20.0168 g For delta t of water: Trial 1 is 15.5 C Trial 2 is 15.7 C Trial 3 is 15.1 C For delta t of metal Trial 1 is 80.1 C Trial 2 is 80.2 C Trial 3 is 79.5 C For B my calculated Specific heat is: Trial 1 is 0.462 Trial 2 is 0.467 Trial 3 is 0.466