Questions/Graphs I need help with Data for graphs Helpful info to answer the above questions Intro I Procedure Prelab (done) From textbook Answer to prelab T1/2 = A0 / 2k T1/2 = Ln2 / k T1/2 = 1 / kA0 A 70 kg patient is treated with ciprofloxacin (2 x 500mg/day) for infection pulmonary Ciprofloxacin is an antibiotic eliminated mainly by the kidneys, its sound apparent distribution volume is 3 1/kg) and its clearance is 05 (Wh x kg) Calculate the halflife time of ofloxacin in thisDear Student, In a response at first sum is 100 % t1/2 implies the time in which it ends up noticeably half then half is remained Next time half of that half disintegrates along these lines, another t1/2 goes That is called t3/4 In this way, 2 t1/2 = t3/4 Here t1/2 = ln2/k
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T1/2 ln2/lambda- t1/2 = ln2 / k t1/2 = 0693 / t1/2 = 7372 days ardni313 ardni313 The halflife of the radioisotope 7368 days Further explanation The core reaction is a reaction that causes changes in the core structure 10%/90%=e^ k*1800 ln10%/90%/1800 = k = 395E4 t1/2= ln(2)/k t1/2 = 693/395E4= 1754s 1754s*1min/60s = 29 min
If 6 0 G Of Substance S Decomposes For 36 Minutes The Mass Of Unreacted S Remaining Is Found To Be 0 750g What Is The Half Life Of Thisreaction If It Follows First Order Kinetics
1/2=ek ·t1/2 ln2=k ·t1/2 t1 /2 =ln2/k Halflife does not depend on concentration D\NEWDDRIVE\pha5127_Dose_Opt_I\Homeworks\Homework1\Fall01\anshmwk1doc 4 Time (hr) Drug Graph indicates zero order elimination because drug concentration vs time results in a straight line Zeroorder elimination CC 0 =k ·tLn 2 is called the halflife of an exponential decay, where ?=RC is the time constant in an RC circuit The current in a discharging RC circuit drops by half whenever t increases by t 1/2 For a circuit with R=3 k?Bisection Method (21) 1 Intermediate Value Theorem If f is continuous on a, b and K is a number between f a and f b , then there exists a number c in a, b for which f c K Note that i If f a f b 0, then either f a or f b is less than 0( K), and there exists a number c in a, b for which f c 0 c is a solution of the equation f x 0 ii The condition in i is a sufficient condition, that
Once you have that, you can determine the half life, which is t1/2 = ln2/k multiply that by 3 to get the total time elapsed for 3 half lives After that, use the first order rate law lnA0 lnAt = kt and solve for At Log in or register to post comments Where, k is the rate constant of the reaction t1/2 is the halflife time of the reaction (t1/2 = 16 years) ∴ k = ln2 / (t1/2) = 0693 / (16 years) = 428 x 10⁻⁴ year⁻¹ For firstorder reaction kt = lna / (ax) where, k is the rate constant of the reaction (k = 428 x 10⁻⁴ year⁻¹) t is the time of the reaction (t = t1/2The halflife and the degradation rate constant are related by the equation t1/2 = ln2/k
The rate constant, k, will be equal to k = ln 2 t 1 / 2 so k = d a y − 1 For the reaction to be 50% complete, that will be exactly the halflife of the reaction at 10 days For the reaction to be 60% complete, using the similar equation derived from question 4, we have A t A o = 04 t = 132 d a y sT1/2 = ln2/k or k = 0693/t 1/2 applies where t 1/2 is the halflife and k is the firstorder rate constant, so k = 0693/36 hrs = hr^1 For a firstorder process rate = kA^1 where k is the firstorder rate constant and A is the amount of radioactive material present5g25g125g063g 87/29 =3 8 A 0456mg sample of hydrogen3 was collected
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1/2 −ln 1 2 =kt /2 ln2=kt 1/2 t 1/2 = ln2 k Note t1/2 independent of A o Means that in each interval of duration t1/2 the concentration decreases to half of what it was at the beginning of the interval, throughout the entire course of the reaction Graphical picture of half life 7 The time constant τ Defined simply as τ =1/k t/t 1/2 1The quantity t 1/2 =? Halflife for pseudo 1st order expts Expt 1 (t1/2) A = mol dm–3 Rate = (kB1) A1 Rate = k'A1 B = 0100 mol dm–3 where k' = kB1 pseudo first order ln2 t1/2 = k' ln2 = = kB ln2 k0100
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k = ln(Af/Ai) / t, and thus T1/2 = ln2/k ln(057) / 74 min = /min, and T1/2 = ln2/0076, = 912 minutes You can mentally check this by looking at the original data You have slightly less thanT1/2 = ln2/k = ln2/0210 = 330 sec b) How long does it take to hydrolyze 75% of the sucrose?The equation can easily be remembered from the definition if said in the right way Clearance is the volume of plasma completely cleared of drug per unit time So Cl = V D x k el And rearranging Cl = V D x ln2 / t 1/2 or t 1/2 = V D
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View Test Prep practice exam 2 from CHEM 102 at Drexel University Practice Exam II CHEM 102 Potentially useful data First order halflife t1/2 = ln2 / k At = kt Ao lnAt = ktThe halflife and the degradation rate constant are related by the equation t1/2 = ln2/k t1/2 = ln2/k = 0693/k So for the avidinbiotin complex t1/2= 0693/ = approx 80 days If we compare data from a range of muscarinic M3 antagonist in the table below it is clear that the reason Tiotropium displays long lasting pharmacological effects despite very low trough levels in the clinic is due to the offrate from the M3 receptor
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And C=3 ?F, if the current is 6 mA at t=5 ms, at what time (in ms) will the current be 3 mA? Homework Statement Show using your expression for \\lambda that if at a time t1 the amount is y1, then at a time t1 ##\\lambda## the amount will be ##\\frac{y1}{2}##, no matter what t1 is Homework Equations y = y0 ##e^{kt}## From previous question half life ##\\lambda =ln2/k## The AttemptThus, they are only order of magnitude comparable F = female, K cond = conditional stability constant, k obs
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T1/2 = ln2/k second order half life t1/2 = 1/(kxAo) shelf life length of time the product may safely be stored on the dispensary shelf before significant decomposition occurs, time taken for 10% active drug decomposition zero order shelf life t09 =1/2=ek ·t1/2 ln2=k ·t1/2 t1 /2 =ln2/k Halflife does not depend on concentration The equations did not have to be deduced I just did so to show from where they actually come from Title anshmwk1PDF Author Pat Khan Created Date Wednesday, AM Explanation I don't really understand the equation you mention in your question, for that I would need more information However, it might help you to know that 0693 is the same as ln(2) This is used for calculating half life / the exponential decay constant λ = ln2 T in which T is the half life of an element
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Ln=ktln t1/t2=ln2/k k=ln/t 2 years It undergoes β− decay into yttrium90, with a decay energy of 0546 MeV b How much of a 50g sample of strontium90 will remain after 87 years?The radioactive isotope of lead, Pb9, decays at a rate proportional to the amount present at time t and has a halflife of 33 hours If 1 gram of thisT1/2 = ln2 / K H = 0693 / K H (5) pH Dependence From the equation (5) it is evident how pH affects the overall rate constant for hydrolysis, K H at high or low pH (high OH or H) either base or acid catalysed term is usually dominant, while at pH 7 the neutral processes can often be the most important
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But Κel is also equivalent to ln2 divided by elimination rate halflife t1/2 (Κel=ln2t1/2) Thus, Cltot = ln2 Vd/t1/2 This means, for example, that an increase in total clearance results in a decrease in elimination rate halflife, provided distribution volume is constant75% = 2 half lives = 660 sec 2 In a first order decomposition reaction, 500% of a compounds decomposes in 105 min a) What is the rate constant of the reaction?How many halflives did the sample go through at the end of 87 years?
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Vì phản ứng bậc nhất nên ta có kt1/2 = ln2 = 0,693 ⇒ k = k 27o C = 0, 693 = 1,3910−4 giây1 5000 b Ở 37oC, thay số ta có k 37o C = 0, 693 1 = 6, 9310−4 giây 1000Start studying chem 2 test 1 Learn vocabulary, terms, and more with flashcards, games, and other study toolsK increases as temperature increase in endothermic reactions Pressure effect on Equilibrium increase in pressure/ decrease in volume favors side of reaction with least moles
Oneclass Part A Half Life Equation For First Order Reactions T1 2 0 693k Where T1 2 Is The Half Lif
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Rationale In the onecompartment model following iv administration the mean residence time (MRT) of a drug is always greater than its halflife (t(1/2)) However, following iv administration, drug plasma concentration versus time (t) is best described by a twocompartment model or a two exponential equationC=Ae(alpha t)Be(beta t), where A and B are concentration unitT1/2 is the halflife time of the reaction (t1/2 = 106 minutes) ∴ k = ln2/(t1/2) = 0693/106 = 654 x 10⁻² min⁻¹ For firstorder reaction kt = lna/(ax) t_"1/2"^((2)) = "8525 s" Notice how you're given the halflife (for one temperature), a second temperature, and the activation energy The key to doing this problem is recognizing that the halflife for a firstorder reaction is related to its rate constant the rate constant changes at different temperatures Go here for a derivation of the halflife of a firstorder reaction
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Ln2 / k first order elimination, k= slope x 2303 first order elimination ln C = ln C0 kt Dose= increases t1/2 by decreasing metabolism cardiac failure decreases clearance so increases t1/2 hepatic failure decreases elimination so increases t1/2 renal failureYou know that in every dynamic 1st order reaction (a reaction with a fixed rate of promotion) we can say that half life is when N/N0=1/2 t1/2= (ln (1/2))/k = ln2/k t1/2=0693/k This formula1 k = ln2/t 1/2 =301/14 = 2 ln A = kt lnAo = 0027*300 ln(22) = 081 3091 lA 228 v = kA lnA = 228 A = 979 µM = 0027*979 = 0026 µM
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CL/Vss = K The use of K derived from the ratio of Vss and CL for halflife calculation (ln2/K) can be applied only to onecompartment model >>This is true that K is the elimination rate constant for a 1 compartment model In the case of the 2 compartment model or higher, CL/Vss = kss, the steadystate rate constant kss = 0693/t1/2ss = 0693O T = time • Can solve this equation for a specific time point – the half life At this time point t = t1/2 and N = N0/2 • Substituting into original equation gives o N0/2 = N0ekt1/2 > 2 = ekt1/2 > ln2 = kt1/2 o Therefore t1/2 = ln2/k • This explains why a large halflife corresponds to a small decay constant (and activity) 4 Calculate ages of artefacts from activity ratiosActually, you don't need to know about radioactive decay constants, λ , "k", etc to do halflife calculations However, if you must learn about these in school, then this is the place to learn it Radioactive Decay Constant λ (lambda) If you need to know about the "k" radioactive decay constant, click here, otherwise just stay here λ (lambda) is defined as the natural log of 2
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The halflife is t1/2 = ln2/k The goodness of fit of the decay model for each gene was assessed with the F statistic (ref ), based on the null hypothesis that the data fit a firstorder decay model A bootstrap method was used to calculate confidence intervals for lnC 0 – ln2 = lnC 0 –kt 1/2 ln2 = kt 1/2 t 1/2 = ln2/k where ln2 = 0693 Now what about clearance? Relevant Equations 𝑇1/2 = (𝑙𝑛2) 𝑥 𝑉𝑜𝑙𝑢𝑚𝑒 𝑑𝑒 𝑑𝑖𝑠𝑡𝑟𝑖𝑏𝑢𝑡𝑖𝑜𝑛/ clarence A 70 kg patient is treated with ciprofloxacin (2 x 500 mg / d) for an infection pulmonary Ciprofloxacin is
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First order (Radioactive decay) t1/2 = ln2 / k or Cheggcom Science Chemistry Chemistry questions and answers First order (Radioactive decay) t1/2 = ln2 / k or 0693 / k 1 It takes 293 hours for a radionuclide to reach 70% of its original activityNote—Halflife (T1/2) is calculated as ln2/k obs Note that K therm, K cond, k obs and T1/2 data were not obtained in identical testing, are not an exhaustive listing, and can be highly dependent on conditions;For half life, t1/2 05 = For time t99% to go to 99% completion 001 = t99% = k , 1/2 = t99% In 4605 ln2 = 0693 t1/2 ln2 k 664 For ok (Table 91) a b In a b c = 77 x 1011 X 1011 X 101 13 x 1011 x 104 = —In (fraction remaining) = kt In 2 tin = k In 2 — = yearsI t1/2 = x 25 = = 0540
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