ipl-logo

Compacted Mixture Lab

1523 Words7 Pages

Lab Report #3:
Laboratory Compaction Characteristics of Soil Using Standard Effort (12,400 ft-lbf/ft3 (600 kN-m/m3))

Soil compaction demonstrates how the soil will behave considering factors such as physical and chemical properties, moisture content, method of compaction, amount of compactive effort, and thickness of layer or “lift” being compacted when a compactive effort was added to it. This results to a compaction curve which is a plot of dry density versus moisture content. This was obtained by compacting moist soil with prescribed added amount of water in a compaction mold and getting the moisture content of a subspecimen retrieved from the center of the compacted mixture. The peak of the curve is a point showing the optimum moisture …show more content…

Data Plotted for the Compaction Curve Figure 1. Plot of Dry Density versus Moisture Content (Compaction Curve)
Equation of Compaction Curve: y = -0.0372x^3 + 0.4098x^2 + 35.729x + 840.15
Derivative of Equation: y ' = -0.1116x^2 + 0.8196x + 35.729
Roots of the Derivative: x_1=21.93775487,x_2=-14.59366885
Optimum Water Content and Maximum Dry Density (using x_1): (21.93775487,1428.433507)

Sample Computations: Water Content
W_N=(M_cms-M_cds)/(M_cds-M_c )×100%=M_w/M_s ×100% (1)
Where:
W_N = water content, %
M_cms = mass of container and moist specimen, g
M_cds = mass of container and oven dry specimen, g
M_c = mass of container, g
M_w = mass of water (M_w=M_cms-M_cds), g
M_s = mass of oven dry specimen (M_s=M_cds-M_c), g
Trial 1:
W_N=18/128×100%=14.0625%
Volume of Mold
V=π/4 D^2 h (2)
Where:
V = volume of mold, m3
D = inner diameter of mold, m h = inner height of mold, m
V=π/4 〖0.1015〗^2 0.115=0.000930507 m^3 Moist Density ρ_m=K ((M_t-M_md))/V (3)
Where:
ρ_m = moist density of compacted subspecimen (compaction point), kg/m3
M_t = mass of moist soil in mold and mold, g
M_md = mass of compaction mold,

Show More
Related

  • Eastmamol/540 Week 4

    563 Words  | 3 Pages

    Prelab week 1 Calculations Preparation of 1.5μmol/L mixed low-level standard dilution 150μmol/L × V1=1.5μmol/L × 10ml V1=(1.5μmol/L×10ml)/(150μmol/L)=0.1ml Conversion of milliliters to microliters (0.1ml×1000)μL= 100μL Preparation of 3μmol/L mixed low-level standard dilution 150μmol/L × V1=3μmol/L × 10ml V1=(3μmol/L×10ml)/(150μmol/L)=0.2ml Conversion of milliliters to microliters (0.2ml×1000)μL= 200μL Preparation of 3μmol/L mixed low-level standard dilution 150μmol/L × V1=7.5μmol/L × 10ml V1=(7.5μmol/L×10ml)/(150μmol/L)=0.5ml Conversion of milliliters to microliters (0.5ml×1000)μL= 500μL Preparation of the blank samples The volumetric flask will be filled to the mark with 150μmole/L of stock solution to act as blank (reference). Additional two blanks will

    Read More

  • Unidentified Ufo Lab Report

    1222 Words  | 5 Pages

    The hypothesis made, the density calculated in the experiment will stay the same because the density of the unidentified object will never change, was supported. The results support the hypothesis because in every trial the density always came out to 9g/mL. In trial one the mass was 71.16g, the volume was 8mL, and the density was 8.895g/mL, but when rounded to the proper sig fig came out to 9g/mL. In trial two the mass was 71.12g, the volume was 8mL, and the density was 8.89g/mL, but when rounded to the proper sig fig came out to 9g/mL. In trial three the mass was 71.14g, the volume was 8mL, and the density was 8.8925g/mL, but when rounded to the proper sig fig came out to 9g/mL. When averaged the mass was 71.14g, the volume was 8mL, and the density was 9g/mL. Errors that could have occurred are, not calculating the density correctly, not completely submerging the unidentified object with water in the graduated cylinder to get the volume, not rounding the sig figs correctly when finding the density, not measuring the unidentified object’s mass in grams, not measuring the unidentified object’s volume in milliliters, and not writing the correct units with the proper number or not the correct unit at all.

    Read More

  • The Crucible: Three Boiling Chips

    292 Words  | 2 Pages

    Now, they are ready to start the lab. The empty crucible is weighed by and the weight was recorded. Then the students used a disposable pipet and put close to 4.0 grams of their milk sample in their crucible on the scale. The crucible was then placed back on

    Read More

  • Unknown 6 Essay

    680 Words  | 3 Pages

    Characteristic Property- Test 2- Density Materials: Triple Beam balance, distilled water, graduated cylinder, unknown 6 Procedure: first we found the mass of the empty graduated cylinder and then its mass with the now distilled unknown. After subtracting the mass of the graduated cylinder, we were able to find the volume. For every 1mL=1cm³ so there we had the volume found with the graduated cylinder. We divided the mass by the volume in order to get the density Data: We found that the density of our unknown was 0.76 g/cm3.

    Read More

  • Transpiration Lab

    1834 Words  | 8 Pages

    The purpose of the transpiration lab was to expand and test one’s knowledge on transpiration, and how it is affected by various factors. This was done in an experiment involving five pansy plants. These pansies were placed into five separate environments with different factors labeled as Control, Intense Light, Humidity, Fan, and Dark. The objective of the lab was to determine how and explain why each environment affected the transpiration of their individual pansey. The main scientific principle being explored in this lab was transpiration.

    Read More

  • Magnesium Sulfate Heptahydrate Lab Report

    478 Words  | 2 Pages

    This heating and cooling was repeated until there was very little (less than 0.0010 grams) fluctuation in numbers. Vial one had a start weight of 14.7681 and an end weight of 15.4098, meaning the mass of the water was 0.4658. Vial 2 had a start weight of 14.7451 and an end weight of 15.3833, meaning the mass of the water in this sample was 0.4633. The mass of the water was found by subtracting the mass of the vial with the hydrate (the start weight) from the mass after the final heating (the final weight). To then find the percent water divide the water mass by the hydrate mass and multiply by 100 since the number is a percent.

    Read More

  • Mass And Mole Relationships In A Chemical Reaction

    949 Words  | 4 Pages

    The items that were massed were the evaporating dish, watch glass, and NaCO3. The materials were massed once before and once after being heated in the drying oven. The mass of the evaporating Dish before was 46.57 g; while after being heating was 60.15 g. The mass of the watch glass before was 57.97 g and after was 48.75g. There were two masses taken for the substance NaHCO3- one with the evaporating dish and one without, subtracted out after the lab was concluded. The mass of the substance with the dish was 48.79 g before and 62.33 g after; meanwhile, the mass of the substance without the dish was 2.22 g before and 2.18 g after. The mass of the NaHCO3 had changed after the reaction occurred along with after it was placed on the hot plate and being in the drying oven.

    Read More

  • Two Components Of An Unknown Mixture Lab Report

    760 Words  | 4 Pages

    The purpose of this experiment was to identify the two components of an unknown mixture through diverse experimental techniques such as recrystallization, extraction, melting point, and acid-base reactions. From this, the group to which these two compounds belong to had to be determined. These groups are: Carboxylic Acids, phenols, and neutrals. By determining the melting points of the two unknown compounds, these values were compared to the values of melting points in the chart and the proper compound was selected. For the case of this experiment, the unknown mixture contained, 4-methylbenzoic acid.

    Read More

  • Nt1310 Unit 6 Lab Report

    631 Words  | 3 Pages

    In this lab there were five different stations. For the first station we had to determine an unknown mass and the percent difference. To find the unknown mass we set up the equation Fleft*dleft = Fright*dright. We then substituted in the values (26.05 N * 41cm = 34cm * x N) and solved for Fright to get (320.5g). To determine the percent difference we used the formula Abs[((Value 1 - Value 2) / average of 1 & 2) * 100], substituted the values (Abs[((320.5 - 315.8) /

    Read More

  • Mixture Lab

    1066 Words  | 5 Pages

    During the Mixed Substances lab the objective was to see how many properties of individual substances compare with properties of mixed substances. From here on the words that would help you along the way of the lab for your research is mixture, homogeneous mixture, and last but least heterogeneous mixture. The hypothesis or guess on what would happen when the water is mixed with the cornstarch, is the cornstarch and water will mix and form a new liquid/solid. However, the Investigating Mixtures lab objective was to observe that not all liquids behave the same way when mixed with other liquids. Moreover the hypothesis of this experiment was that the vegetable oil and colored will separate because the vegetable oil will go to the top and the

    Read More

  • Gummy Bear Lab Report

    1183 Words  | 5 Pages

    The final mass could be far off due to the water and chunks of expanded gummy bears found in the beaker, leading to an inaccurate result. As well, for the sugar solution, the result could have been different if a more accurate measurement of the sugar needed was made. For the specific result, the hypothesis stated, the sugar solution needed to have an equal amount of sugar content to the gummy bear which did not occur. Ensuring that the beaker contained 10 grams of sugar was off, due to prerequisite calculations that lead to too much liquid in the beaker that needed to be removed. To be correct, the hypotheses that were wrong could

    Read More

  • Dust Bowl Cons

    1114 Words  | 5 Pages

    In 1935, Congress heeded to the public’s consternations when it passed the Soil Conservation Act, entrusting $2 million for soil preservation programs, after realizing the Dust Bowl’s threat on the nation’s environment (Yancey 71). The Soil Conservation Act developed the Soil Erosion Services (SES) which bestowed farmers with the equipment and seeds that they needed. Through research at Dalhart, Texas, SES scientists discovered by removing livestock from crop residue and refraining from burning it lessened wind erosion because the remains assisted in keeping the soil in tact. As a result of further scientific revelations, many began planting trees in columns to serve as windbreakers (“Dust Bowl” 5). Subsequently, the Soil Conservation Services was formed to assert a prevention program on soil erosion, teaching farmers proper techniques in which to avoid soil erosion (Loring 1-2).

    Read More

  • Lab Report: Diffusion And Osmosis

    1274 Words  | 6 Pages

    We zeroed out the scale and weighed all four potato cores at once and recorded the mass. We then put those potato cores into the beaker of 75 mL of solution. With the potato cores in the beaker we then put a watch glass over the top of the beaker to minimize the amount of solution that evaporates. We let the potato cores sit in the solution overnight. The next day we then emptied the beaker of the solution by carefully draining the solution, while not letting the potato cores fall out.

    Read More

  • Wittig Reaction Lab Report

    1125 Words  | 5 Pages

    The final product weight for percent yield was only the solid E product, which missed one half of the final product produce. If both products were weight, the percent yield would have been larger that it was. Instead of 22.33%, it could have been 44.66%. To prove that both products were obtained, but only one of the two products was analyze, a TLC plate of the DCM layer, that contains both products, and of the final product, was obtain.

    Read More

  • Stoichiometry And Limiting Reagents Lab Report

    1237 Words  | 5 Pages

    (0.01 moles of NaOH) x (1 mole Ca(OH)2/ 2 moles of NaOH) = 0.005 moles of Ca(OH)2 Tube 1: (0.0020 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.002 moles of Ca(OH)2 (0.002 moles of Ca(OH)2) x (74.08 grams/mole) = 0.1 grams = theoretical yield Tube 2: (0.0035 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.004 moles of Ca(OH)2 (0.004 moles of Ca(OH)2) x (74.08 grams/mole) = 0.3 grams= theoretical yield Tube 3 (0.0050 moles of CaCl2) x (1 mole Ca(OH)2/ 1 mole of CaCl2) = 0.005 moles of Ca(OH)2 (0.005 moles of Ca(OH)2) x (74.08 grams/mole) = 0.4 grams =theoretical yield Tube

    Read More

More about Compacted Mixture Lab

Open Document