The first thing I did was multiplied all the terms by H^ in order to remove the denominator. I then divided all the terms by 703 to isolate the W. My height was then added in order to find out the second weight interval (4096). Multiplication and division were then performed. Those of people whom may weigh between 134lbs and 145.6lbs are most likely not over
We made this inference due to our slope of each of the graphs. Given the mole ratios, the slope of our newer lab, comparing H2 to moles of Mg, should’ve been a slope of 1. But our first point skewed our results giving us an actual slop of 0.7, which was short from a more accurate result. And given our results last week, we concluded that our titration method was more accurate than our crystallization method due to the basically perfect slope of 2.0018 which is very close to the theoretical slope of 2. 5.
10/4/2011 VHS Block 2 Shoe Size and Height Questions 1. While doing this study, there could have been a couple of experimental errors. I could have accidently not produced a 100% accurate measurement for my shoe size, foot size, or height. The reasoning behind this is because since the ruler was not flexible to bend to the shape of my foot or rounded the end of my toe and heal, I had to eye ball the exact measurement. This being said, there was definitely room for some sort, although small, measurement reading error.
Carry it over to the right. 32 is less than 63 so you will subtract and put a 1 for that value, and continue down the line. 2. Explain why the values 102 and 00102 are the same. Because when you plug the binary numbers into the 8 bit conversion table, the two zeros before the 10 equal nothing.
We moved around the weights until we ended up with what we saw to be an even leveled meter. We repeated this a second time using two slightly different weights and recorded them. Our third trial was done in a slightly different way, instead of having a single weight on each side, we placed two masses on the right side of the meter stick and only one on the left. Again, we played with the weights until we were at equilibrium and recorded our data. The second half of the experiment was done in a similar fashion, except the ruler was no longer hung on
Our scale only measures to grams accurately, and could have not accurately represented the .68g of CaCO3, or we could have lost the solid precipitate in the process of the experiment. These errors could be reduced with a more accurate scale, as well as more experienced hands performing the procedure, or skilled hands, ie: we are novices at this lab, and training could produce more accurate
Experiment 5- Limiting and Excess Reagents Purpose The purpose of this experiment was to calculate product and reactant amounts using reaction stoichiometry, as well as calculating the limiting and excess reagent in a copper (II) nitrate and potassium iodide reaction. We tested the reaction with different amounts of Cu(NO3)2 and KI making duplicates of each, with a total of 4 test tubes per group. Procedure Found in lab manual starting on page 91. 1. Collect 4 small test tubes and weight them.
The uncertainty in the distance is ±0.5 cm because the ruler had 1 cm markings. The uncertainty in the time was calculated by using the equation (max time – min time)/2. Occasionally the bearing hit the side of the tube on it’s descent , this would decrease it’s velocity , when this happened consistently throughout one trial the results were ignored . If the bearing
All stones exposed to water did not loose any mass. Since Vinegar is a simulator of acid rain we can predict that Granite and Sandstone should be used. I could have gotten better data if I used a more precise scale. My scale only measured to the half of a gram so it was hard to find a change in mass. Also I could have used the same shape of stone because surface area could have been a
Then finally I divide all 3 by 703 in order to isolate W. 128.8 < W < 166.7 So people with the height of 73 inches could have a longer than average life span if they weigh between 129 and 167 pounds. The next equation 23<703WH^2<25 I first multiply all 3 terms by H^2 in order to take our the denominator. 23H^2 < 703W < 25H^2 Then I divide all terms by 703 to isolate the W. 23H^2703<W< 25H^2703 After doing this it now becomes an equivalent inequality for me to solve the second weight. I now use my height squared to find the second weight interval. 23(5329)703<W< 25(5329)703 After I multiply I end up with 122567703<W< 133225703 Then I simply divide and end up with 174 < W < 189.5.