Q: Isotopes are defined as atoms ofa. the same element with different numbers of protons.b. radioactive…
A: Atoms of the same element have the same number of protons, which is also called the atomic number.
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A: Interpretation: What is radioactive iodine is to be determined.
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A: Opposite charges attract each other while similar charges repel each other. Therefore, electrons and…
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A: NOTE : Since you've posted a question with multiple sub-parts,we'll solve first three sub-parts for…
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Q: True or False? Atomic weight is equal to the number of protons and neutrons in the nucleus.
A: Given : Statement "Atomic weight is equal to the number of protons and neutrons in the nucleus"
Q: Which particle, if lost from the nucleus, will not result in a change in the atomic mass number?
A: Which particle, if lost from the nucleus, will not result in a change in the atomic mass number?
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A: As per bartleby guidelines I answered only first question so please don't mind.Thanks in advance.
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A: Given that, Atomic number of U is 92.
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A: The various names of radioactive elements are named as in step 2.
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Q: Describe Rutherford’s nuclear theory?
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Q: 1. How do atoms attain stability?
A: Since you are posted with multiple questions. As per the rule, I am answering the first question…
Q: What did the scientists who are credited with discovering the proton and neutron have in common?
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Q: An 18 O atom contains the same number of protons. neutrons, and electrons.
A: Given atom, 8O18
Q: In 1909, a team of British scientists led by Ernest Rutherford, carried out the Gold Foil Experiment…
A: Gold Foil Experiment helped to determine the arrangement of particles in the atom.
Q: Complete the table below to compare protons, neutrons, and electrons in terms of their location and…
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A:
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- Use this video to answer the questions: https://www.youtube.com/watch?v=BocfSXdbaU0You are trying to come up with a drug to inhibit the activity of an enzyme thought to have a role in liver disease. In the laboratory the enzyme was shown to have a Km of 1.0 x 10-6 M and Vmax of 0.1 micromoles/min.mg measured at room temperature. You developed an uncompetitive inhibitor. In the presence of 5.0 x 10-5 M inhibitor, the apparent Vmax was determined to be 0.02 micromoles/min.mg. What is the Ki of the inhibitor?Using data in Appendix E, calculate the standard emf foreach of the following reactions:(a) H21g2 + F21g2 ¡ 2 H+1aq2 + 2 F-1aq2(b) Cu2+1aq2 + Ca1s2 ¡ Cu1s2 + Ca2+1aq2(c) 3 Fe2+1aq2 ¡ Fe1s2 + 2 Fe3+1aq2(d) 2 ClO3-1aq2 + 10 Br-1aq2 + 12 H+1aq2 ¡ Cl21g2 +5 Br21l2 + 6 H2O1l2
- 5.92 x 10^-15 ksp according to aleksAn enzyme-catalyzed reaction has a Km of 1.4 mM and a Vmax of 7 nM/s. What is the initial velocity when the substrate concentration is 0.6 mM?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…
- 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=drivesdk0% 25% 50% 75% 100% Depth of H2O2 Solution (d) 2.1 cm 2.1 cm 2.1 cm 2.1 cm 2.1 cm Trial 1 Time 180 sec 84.61 sec 43.52 sec 36.90sec 25.90 sec Trial 2 Time 180 sec 92.25 sec 38.16 sec 34.36 sec 23.57sec Trial 3 Time 32.53 sec 18.82 sec Average Time (t) 180 sec 88.43 sec 40.84 Sec 34.5 sec 22.76 sec Rate of the Reaction(R = d/t) 0.012 cm/sec 0.02 cm/sec 0.051 cm/sec 0.061 cm/sec 0.0923cm/sec , graph rate of reaction on the y-axis and percent concentration of enzyme on the x-axis. If the points are linear, draw a “best-fit” straight line through or near all of the data points. Based on the information in the data table and your graph, explain the relationship between percent concentration of catalase and rate of reaction. Did your actual results match your hypotheses? If not, why?https://m.youtube.com/watch?v=vM1SP346XBc&list=PLeJOSNLNZfHubfLdq0kOayASeUllMOGn4&index=4 I watched this the lecture video over and over and I am allowed to work with someone but I am having trouble with part B and I provided the YouTube link of the data or video attached to this lab
- PLS HELP VERY URGENT Q20a) Determine kcat (in units of sec-1) for a particular enzyme, given the following information: Vo = 144 mmol/min; [S] = 2 mM; Km = 0.5 mM; Enzyme Molecular weight = 40,000 micrograms/micromole; 8 micrograms of enzyme used in assay generating this data. b) In general, explain how the total enzyme concentration affects turnover number and Vmax?Please create a caption for this table. Solution NaCl Conc. (%) Osmolality (mOsm) % transmittance Absorbance % hemolysis % crenation C distilled 0 0 0.001029 4.987584625 100 0.03354 1 0.177179111 54.61 0.001551 4.809388202 96.42720001 0.05837 2 0.297126222 91.58 0.01012 3.994819487 80.09527231 0.08444 3 0.442542222 136.4 3.849 1.414652089 28.3634704 0.134 4 0.590164444 181.9 64.8 0.188424994 3.777880643 0.2125 5 0.74752 230.4 95.64 0.019360433 0.388172513 0.3368 6 0.89644 276.3 99.56 0.001915112 0.038397585 0.5336 7 1.095648889 337.7 99.98 8.68676E-05 0.001741676 0.9834 8 1.336711111 412 100 0 0 2.1 9 1.755568889 541.1 100 0 0 7.9 10 2.674395556 824.3 100 0 0 57.83 11 4.490211111 1384 100 0 0 99.72