<span>Ores are naturally occurring rocks that contain metal or metal compounds in sufficient amounts to make it worthwhile extracting them. For example, iron ore is used to make iron and steel.</span>
Answer:
A. percentage mass of iron = 5.17%
percentage mass of sand = 8.62%
percentage mass of water = 86.205%
B. (Iron + sand + water) -------> ( iron + sand) ------> sand
C. The step of separation of iron and sand
Explanation:
A. Percentage mass of the mixtures:
Total mass of mixture = (15.0 + 25.0 + 250.0) g =290.0 g
percentage mass of iron = 15/290 * 100% = 5.17%
percentage mass of sand = 25/290 * 100% = 8.62%
percentage mass of water = 250/290 * 100% = 86.205%
B. Flow chart of separation procedure
(Iron + sand + water) -------> separation by filtration using filter paper and funnel to remove water --------> ( iron + sand) -----------> separation using magnet to remove iron ------> sand
C. The step of separation of iron and sand by magnetization of iron will have the highest amount of error because during the process, some iron particles may not readily be attracted to the magnet as they may have become interlaced in-between sand grains. Also, some sand particle may also be attracted to the magnet as they are are borne on iron particles.
Answer:Chemistry problems can be solved using a variety of techniques.
Explanation: Many chemistry teachers and most introductory chemistry texts illustrate problem solutions using the factor-label method. ... The use of analogies and schematic diagrams results in higher achievement on problems involving moles, stoichiometry, and molarity. Hope this helped!
Answer:
4.94g of material
Explanation:
Partition coefficient (Kp) of a substance is defined as the ratio between concentration of organic solution and aqueous solution, that is:
Kp = <em>8 = Concentration in Ethyl acetate / Concentration in water</em>
100mL of a 5% solution contains 5g of material in 100mL of water. Thus:
8 = X / 100mL / (5g-X) / 100mL
<em>Where X is the amount of material in grams that comes to the organic phase.</em>
8 = X / 100mL / (5g-X) / 100mL
8 = 100X / (500-100X)
4000 - 800X = 100X
4000 = 900X
4.44g = X
<em>Thus, in the first extraction you will lost 4.44g of material from the aqueous phase.</em>
And will remain 5g-4.44g = 0.56g.
In the second extraction:
8 = X / 100mL / (0.56g-X) / 100mL
8 = 100X / (56-100X)
448 - 800X = 100X
448 = 900X
0.50g = X
<em>In the second extraction, you will extract 0.50g of material</em>
Thus, after the two extraction you will lost:
4.44g + 0.50g = <em>4.94g of material</em>
<em></em>