Wreath Problem
The goal of the wreath problem was to find out how much gold was truly found within the wreath. The materials we used for this lab included a scale, a wreath, a beaker, a “gold” object, and a “silver” object. After getting our materials, we weighed each object, the gold, silver, and wreath, on the scale in order to find how much each object weighed in grams in air. The gold weighed 56.3125 grams. The silver weighed 35.19 grams. The wreath weighed 169.99 grams. After measuring the weight of each object in air, we measured the objects in water in order to find the density for each. The gold weighed 50 grams/milliliters. The silver weighed 30.26 grams/milliliters. The wreath weighed 149.78 grams/milliliters. Once we found the density of each object, we then sought to find the amount of fluid displaced
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To do this, we subtracted the weight of each object in air by the amount the object weighed in water. Therefore, the amount of fluid displaced for the gold object was 6.3125 grams/milliliters. The amount of fluid displaced for the silver object was 4.93 grams/milliliters. The amount of fluid displaced for the wreath was 20.21 grams/milliliters (F). Afterwards, we attempted to find out how much gold was truly found in the wreath. To do this, we used the algebraic method, but modified it a bit. It was stated that W is to equal the weight of the wreath, which also equaled w1+w2, where both w1+w2 represented the weights of gold and silver in the wreath. We then had to solve for W of pure gold. To do this, we set up a ratio. The ratio was as follows: weight of object in air: weight of wreath in air:: weight of the object in water: weight of pure wreath in water. The gold object and our wreath were substituted in the ratio. Therefore, the ratio was 56.3125:169.99:: 50:X. We then rewrote the ratio to look as follows,