Five different fertilizers were analyzed by two methods. Phosphate (P:Os) was analyzed using gravimetric analyse and spectrophotometric method. Analyze the results of the two methods and answer the following questions. Sample Gravimetric result Spectrophotometric results 1 25.5% 24.4% 2 9.2% 10.0% 26.2% 25.8% 4 50.5% 47.3% 5 25.6% 28.6% a Calculate the average content of P:Os in the fertilizers for each method. Compare the difference between the two average values. Im 3
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- Must answer all questions eslse downvote A. MULTIPLE CHOICE. Choose the BEST answer. PLEASE HELP ME ANSWER EVERYTHING THANK YOU Q)What is used for heating small amounts of solids at a high temperature?a) Mortar and pestleb) Evaporating dishc) Crucible and coverd) Clay triangleQ)Which of the following is used in separation techniques?a-Rubber policemanb-Graduated cylinderc-Volumetric flaskd-Filter paperQ)Which of the following describes a centrifugate?a-Always clearb-Supernatant liquidc-Discarded via decantation onlyd-Solid particlesQThe inward force that pulls substances towards its center is called .a-Gravitational forceb-Centrifotal shiftc-Continental shiftd-Centrifugal forceQA "slippery floor" is considered a .a-hazardb-riskc-flash pointd-toxicantQ)Mrs. Lily Potter is 24 weeks pregnant. She was exposed to a chemical while making a potion. What is the type of the chemical she should be avoiding?a-neurotoxicantb-asphyxiantc-teratogend-carcinogenQ)Which of the following should be…A certified reference material of lead in calcium carbonate is analyzed by a laboratory and found to have a lead content of 237 ng/g.237 ng/g. The reported certified value for this material is 211 ng/g.211 ng/g. What is the absolute error and the percent relative error for the method used in this analysis? absolute error: percent relative error:A clinic had 25 patients on Friday morning. If 21 patientswere given flu shots, what percentage of the patientsreceived flu shots? Express your answer to the ones place.
- A student obtained 2.111 g of a mixture containing four components. After performing separation by chemical and physical methods, the student obtained the following mass data: Component 1: 0.486 g Component 2: 0.317 g Component 3: 0.127 g Component 4: 0.929 g Calculate the total percent recovery of the mixture. (Please show work)A titrimetric method for the determination of calcium in limestone was tested with the analysis of a NIST limestone containing 30.15% CaO. The mean of the four analyzes is 30.26% CaO with a standard deviation of 0.085%. From the data accumulated from many analyzes, s→ϭ=0.094% CaO was found.a) Do the data indicate the presence of systematic error at the 95% confidence level?b) When a value for ϭ is unknown, do the data show a systematic error at the 95% confidence level?You work in a research organization that is looking for markers of various diseases that can be used as a diagnostic for the disease. It has been reported in the past that high levels of Cu are found in the sweat of people with cystic fibrosis. One of the research projects is focused on looking for high levels of Cu in samples that can be obtained non-invasively such as saliva, sweat, hair, nails, etc. The lab will analyze large samples for Cu. What instrument would you recommend purchasing to support this work, Atomic absorption spectrophotometer or an inductively coupled plasma atomic spectrophotometer? Explain the basis for your decision.
- You take a representative soil sample and determine the following data: Exchangeable cation cmolc/kg of soil Ca2+ 2.7 Mg2+ 1.0 K+ 0.1 H+ 1.0 Al3+ 2.8 Na+ 0.1 1) What is the CEC (cmolc/kg ) of this soil?Please answer fast it’s very important and urgent I say very urgent so please answer super super fast please 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 0.466Chemistry The levels of an organic pollutant (P) in the groundwater at the perimeter of a plant were a cause for concern. A 10 mL sample of the water was taken and the pollutant was extracted with 95% efficiency using 25 mL of diethyl ether. GC was used to analyse the concentration of P in diethyl ether. A calibration curve was plotted for a series of standards of P which yielded the following results: Peak Area Toluene Conc. (µg/ml) 12,000 2.6 23,700 5.0 35,500 7.7 46,800 9.9 31,250 Sample Determine the concentration of P in ppb in the initial groundwater sample.
- 1. Which of the following is NOT a true representation of the Lineweaver-Burk equation? Group of answer choices a. y-intercept = -1/Vmax b. x-intercept = -1/Km c. y = 1/V0 d. x = 1/[S] e. slope = Km/Vmax 2. If the y-intercept of a Lineweaver-Burk (double reciprocal) plot = 39.37 (sec/millimole) and the slope = 75.3 L/sec, Vmax equals: Group of answer choices a. 75.3 millimoles/sec. b. 0.0254 millimoles/sec. c. 0.523 millimoles/sec. d. 39.4 millimoles/sec. 3. Which of the following is true regarding allosteric enzyme behavior?Group of answer choices a. Allosteric enzymes are rarely important in the regulation of metabolic pathways. b. An allosteric effector binds reversibly on the surface of the enzyme. c. An hyperbolic representation of the relation initial velocity vs [S] is a specific characteristic of an allosteric enzyme behavior. d. Michaelis-Menten kinetics model the kinetic behavior of allosteric enzymes.Given the Ksp for Ag2CrO4 is equal to 1.20 x 10-12, calculate the percent error in your experimentally determined value for average Ksp (this will be an extremely large percentage – think about why).A water sample was analyzed for Fe content using the iron-phenanthroline method. The following data were obtained from the analysis:(photo) Determine the value of the slope (m), y-intercept (b), and the coefficient of determination (r2) m = 0.1128; b = -4.0 x 10-5; r2 = 0.9880 m = 0.1128; b = - 3.0 x 10-3; r2 = 0.9762 m = 0.1128; b = - 3.0 x 10-3; r2 = 0.9880 m = 0.1128; b = -4.0 x 10-5; r2 = 0.9762