The correct answer is From different social and ethnic backgrounds.
Explanation:
The purpose of science studies is to gain an understanding of specific phenomena and use this in the benefit of the world or humanity. Moreover, science is guided by objective methods and these are applied all around the world. This feature allows scientists from all kinds of social and ethnic backgrounds to study the same phenomena and obtain similar results, which is essential for the progress of scientific knowledge. This is exemplified by two different labs in two different countries studying drug absorption because the results of these two studies can be analyzed together to understand drug absorption.
In this context, one factor that contributes to the progress of scientific knowledge is the effort of scientists "from different social and ethnic backgrounds".
Answer:
Final molarity of iodide ion C(I-) = 0.0143M
Explanation:
n = (m(FeI(2)))/(M(FeI(2))
Molar mass of FeI(3) = 55.85+(127 x 2) = 309.85g/mol
So n = 0.981/309.85 = 0.0031 mol
V(solution) = 150mL = 0.15L
C(AgNO3) = 35mM = 0.035M = 0.035m/L
n(AgNO3) = C(AgNO3) x V(solution)
= 0.035 x 0.15 = 0.00525 mol
(AgNO3) + FeI(3) = AgI(3) + FeNO3
So, n(FeI(3)) excess = 0.00525 - 0.0031 = 0.00215mol
C(I-) = C(FeI(3)) = [n(FeI(3)) excess]/ [V(solution)] = 0.00215/0.15 = 0.0143mol/L or 0.0143M
a. ELEMENT: A substance<span> that </span>cannot<span> be decomposed (</span>broken down<span>) </span>into simpler substances by ordinary chemical<span> means.</span>
It is B, From the bottom of the pyramid to the top
Answer:
1.30464 grams of glucose was present in 100.0 mL of final solution.
Explanation:

Moles of glucose = 
Volume of the solution = 100 mL = 0.1 L (1 mL = 0.001 L)
Molarity of the solution = 
A 30.0 mL sample of above glucose solution was diluted to 0.500 L:
Molarity of the solution before dilution = 
Volume of the solution taken = 
Molarity of the solution after dilution = 
Volume of the solution after dilution= 



Mass glucose are in 100.0 mL of the 0.07248 mol/L glucose solution:
Volume of solution = 100.0 mL = 0.1 L

Moles of glucose = 
Mass of 0.007248 moles of glucose :
0.007248 mol × 180 g/mol = 1.30464 grams
1.30464 grams of glucose was present in 100.0 mL of final solution.