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- Calculate the following exercise with correct sig figs: ((2.9365mL - 3.00mL)/3.00mL)x100%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 âdriftâ of the resistance readings?C.What is the equation of your best-fit line?D.What commercial drink did you analyze?E.Assuming…NH4+ {aq) + NO2(aq) -> N2(g) +2H2O{l} Data Initial [NH4+] Initial [NO2-] rate 1 0.0100 0.200 5.4 x10-7 2 0.0200 0.200 10.8x10-7 3 0.0400 0.200 21.5x10-7 4 0.200 0.0202 10.8x10-7 5 0.200 0.0404 21.6x10-7 6 0.200 0.0808 43.3x10-7 Find x,y,k
- pls help thermodynamic chemComputers are not supposed to be in very warm rooms. The highest temperature tolerated for maximum performance is 308 K. Express this temperature in C and F.Hiii!! does anyone here can help me with this said activity? How can i know what the vid is telling ( https://youtu.be/dinNjEiqxcg , https://youtu.be/Xj0X3CS9J3k , and https://youtu.be/1IaaMeGQwdg ). thank u