The results of a field unit weight determination test performed by means of the sand cone method are as follows: Calibrated dry unit weight of Ottawa sand = 104 pcf Weight of Ottawa sand to fill the cone = 0.258 lb Weight of jar + cone + sand (before use) = 13.21 lb Weight of jar + cone + sand (after use) = 6.2 lb Weight of moist soil from hole = 7.3 lb Moisture content of moist soil = 11.6% Determine the dry unit weight of compaction in the field. 0 0 0 0 0 0

A clean sandstone core sample with diameter of 3.8 cm and length of 5.6 cm is saturated with a brine. The core was desaturated in several steps, and the resistance was measured at each saturation. The results are as follows: Water Saturation (Fraction) Resistance (kS) 1.000 1.302 0.900 1.695 0.800 2.283 0.730 2.878 0.640 3.775 0.560 5.637 0.480 7.838 0.375 13.175 0.350 17.050 0.300 26 Brine resistivity= 0.74 2m Archie coefficient 1.03 Cementation factor= 2.2147 (1) Estimate the rock porosity. (2) Determine the saturation exponent. (3) Estimate the amount of oil and brine (in cc) in the core sample when resistivity index is 5. All steps of the calculations should be shown.

SG is 2.66 and temperature is 18 degrees celcius

A liquid limit test on a clay was performed with the following results. The natural water content of the clay is 38% and plastic limit is 21%.   Number of Blows 6 12 20 28 32 Water content (%) 52.5 47.1 42.3 38.6 37.5   What is the plasticity index of this clay? (Use Interpolation) Group of answer choices 15.74% 18.99% 18.85% 17.63%   Please answer this asap for upvote. thanks