1. Larry, who has 20% hct and 95% oxygen saturation would benefit by a transfusion because his hct is low, but does not need supplemental oxygen. Curly who has 40% hct and 95% oxygen saturation would be helped by neither because both hct and oxygen saturation percentage is in the normal range. Moe who has 30% hct and 95% oxygen saturation would benefit from a transfusion because his hct is low; however, supplemental oxygen would not be of any assistance.
2. a. At 60 mm Hg, the data values in the table are not consistent with the data in the plots; for example, total mm Hg O2 bound is 30 with oxygen for Bonnie and 2 for Clyde. On the conventional plot, the 60 mm Hg is depicted at 50 with oxygen for Bonnie and about 9 for Clyde. The table says Bonnie should be shown at 6 instead of 19 at this concentration on the seating plot; also, the table says Clyde should be shown at 0 additional amount bound instead of 10 on the seating plot.
b. On the seating plot, Bonnies
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I started looking at this problem by looking at the table, as I checked the math of the table I noticed some errors. I rewrote the table on a piece of notebook paper and I saw where some gaps were. The numbers in red are the ones that I calculated to be different than the ones provided. Next I re-drew each plot, both conventional and the seating plot to better understand where the graphs varied. I noticed that around 60 mm Hg in most graphs had inconsistences so I decided to take my time and understand those points, next I looked at the plots on the exam and looked at how each plot differed from mine. I also used the conventional plot to analyze the saturation change, like the way I learned in lab 12 regarding the question of Gale Bladder and Baby Sue. The approach that best worked for me is the table of values because it is simple, and straight forward. The table does not involve any interpretation; I rather have exact numbers than attempting to guess the numbers on the seating and conventional
INTRODUCTION: This case involves the suspect being arrested for public intoxication and unable to care for himself in violation of PC 647(f)-Public Intoxication. The suspect was later placed on a WIC 5150-Mental Health Hold. INVESTIGATION:
Question3: Experiment 3 The unknown acid sample was 1 • Monoprotic Acid Trails Initial NaOH solution (mL) final NaOH solution (mL) The volume of NaOH to titrate the acid (mL) Amount of Unknown Acid sample 1 (g) The moles of the Unknown Acid (mol) Molar mass of the Unknown Acid (g/mol) A 3.38 28.31 24.93 0.150 0.0026 57.69 B 0.18 29.32 29.14 0.175 0.0029
75 90 Doc 1 1.31556 1.80357 1.80357 Doc 2 1.75182 1.87012 1.87012 Doc 3 2.13338 2.27178 2.27178 Doc 4 1.56941 1.0743 1.0867
Apply 5.2 Display quantitative data with appropriate descriptive statistics (mean, SE) on a graph. Paste your graph here. 5) Apply 5.3: Display quantitative data with p-values for differences between means. Apply 5.4: Understand what statistical differences between means indicate. Report your plankton means ± 2SE with the p-value for comparing those means.
However, during his workout he was slightly dehydrated since his urine turned to dark yellow. The urine color and specific gravity are matched in terms of normal values based on the data results. Because the normal specific gravity of urine ranges from 1.000-1.030 according to medlineplus.gov. 2. Based on the urine color and specific gravity, what might Tracey conclude about the hydration status of Max 's body at the three different times? Answer: Since
observations ½ tab 17.9 55.30 1 ½ tab divided in 2 pc. 19.5 49.37 The pieces of alka-seltzer were drawn to each other in the cup ½ tab divided n 3 pc. 20.8 37.24 The pieces of alka-seltzer were drawn to each other in the cup ½ tab 19.4 52.01
His symptoms are linked to his CBC results that are below the reference range of RBC, hematocrit and hemoglobin. Harold has an elevated mean of RBC volume while having a low RBC and hematocrit count, this indicates that he is producing an abnormally large RBC that does not efficiently carry oxygen. 3. Doctors have determined that Harold is anemic. Describe this condition.
Materials -pan -50ml graduated cylinder -hydrogen peroxide -air stopper -water Graphs data A time 12 drops 8 drops 0 0 0 30 0 0.5 60 0 1 90 0 1 120 0 1 150 0 1 180 0 1.5 210 0 1.5 240 0 1.5 270 0 1.5 300 1 1.5 330 2 1.5 360 2 1.5 390 3 1.5 420 3 1.5 450 3 2 480 3 2 510 3 2 540 4 2.5 570 4 2.5 600 5 2.5 Data B time cold warm 0 0 0 30 1 1 60 2 1 90 2 2 120 2 2 150 2 2 180 2 2 210 2.5 2 240 3 2 270 3 2 300 3 2 330 3 3 360 3.5 3 390 3.5 3 420 3.5 3 450 3
Use your results in Data Table 2 to support your answer.
1. List five warning signs for each of the clients in the case studies (10 points) Case #1John 1. Sleep difficulties and increased agitation. 2. Lack of social activity.
In this lab we used two processes called Diffusion and Osmosis. Diffusion is the movement of molecules from areas of high concentration to areas of low concentration. Diffusion is a process that requires no energy and involves smaller non-polar molecules. In Figure 1 you can see the molecules spreading throughout the glass from the area of high concentration, so that the areas with low concentration are filled evenly as well. The other process was osmosis.
Predict/ roughly determine the Vmax and ½ Vmax values from the peak of the graph, where the slope of the graph levels off (the asymptotical line). Predict/ roughly determine the Km by reading off of the graph the corresponding substrate concentration on the x-axis for the ½ Vmax value. Plot a Lineweaver-Burke graph (the inverse of the velocity of the reaction vs. the inverse of the substrate concentration). Calculate accurate Vmax and Km values using the following equation for the Lineweaver-Burk
When R2 equals 1.0, all points lie exactly on a straight line with no scatter. Knowing X lets you predict Y perfectly. (http://www.graphpad.com/guides/prism/6/curve-fitting/index.htm?r2_ameasureofgoodness_of_fitoflinearregression.htm) In this graph the R2 value is 0.97 which can be rounded off to 1.0 which means that knowing X which in this case is the different concentrations of sodium thiosulfate, predicts the Y which is the time the time the solution turns cloudy resulting X not to be seen from the opening of the conical flask from a person’s eye
IV. Data and observations Mass of beaker (g) 174.01 Mass of beaker + NaOH pellets (g) 174.54 Mass of NaOH pellets 0.53 TRIAL 1 TRIAL 2 Mass of potassium acid phtalate (KHP) (g) 0.15 0.15 final buret reading (ml) 30.75