Answer:
D) Repeating the experiment several times and comparing the results
Explanation:
Replication of experiments allows scientists to see patterns and trends in the results of the experiments. It helps the scientist performing the experiment to see if there is precision in the results gotten and to test the integrity of the data. The results of several repetitions of the experiment can be compared and tested for consistency.
Performing this experiment severally will make the scientist know whether the dissolved oxygen contents gotten in each of the tanks are close to each other. If the results are wide apart, then the data lacks integrity and should not be used.
Answer:
It shows the same number of atoms of each element on both sides of the equation
A balanced equation demonstrates the conservation of mass by having the same number of each type of atom on both sides of the arrow.
Explanation:
Matter cannot be created or destroyed in chemical reactions. This is the law of conservation of mass. In every chemical reaction, the same mass of matter must end up in the products as started in the reactants. Balanced chemical equations show that mass is conserved in chemical reactions.
Answer:
A) refraction experiment n = n₁ sin θ₁ / sin θ₂
B) n_A = 1.19
, n_B = 1.53
Explanation:
A) This exercise is a method to measure the refractive index of materials, a very useful and simple procedure is to create a plate of known thickness from each material, place the material on a paper with angle measurements (protractor), incline the laser beam and measure the angles of incidence and refraction (within the material), repeat for about three different angles of incidence and use the equation of refraction to determine the index
n₁ sin θ₁ = n₂ sinθ₂
n₂ = n₁ sin θ₁₁ /sin θ₂
If the medium surrounding the plate is air, its refractive index is n₁ = 1, the final expression is
n = n₁ sin θ₁ / sin θ₂
B) For this part, no data are given in the exercise, but we can take 50º as the angle of incidence and measure the angle of refraction. Suppose it is 40º for material A and 30º for material B, the refractive index would be
material A
n_A = sin 50 / sin 40
n_A = 1.19
material B
n_B = sin 50 / sin30
n_B = 1.53
Placing elements into columns, groups, rows and periods that share certain properties. properties such as gas, solid and liquid determine an element physical state at room temperature.
