Rare earth elements are a series of chemical elements found in the earth's crust and are vital to many of the modern technologies in the world such as computers and networks, advanced transportation and consumer electronics. They help fuel economic growth, maintain high living standards and even save lives. Examples include:
Scandium. Used in television and fluorescent lamps.
Yttrium. Used in cancer treatment drugs, superconductors and camera lenses
Lanthanum. Used to make special optical glasses, telescope lenses and also in petroleum refining.
Neodymium. Used in making some of the strongest permanent magnets, found in most modern vehicles and aircraft.
Pressure on the inside of the balloon was greater than the pressure on the outside of the balloon so it pushed out until the pressures equalized.
Option C
Plants, cellular respiration accurately represents a carbon source and the process that releases carbon from that source
<u>Explanation:</u>
Every existing body supplies CO2 off while they evoke power from their food through cellular respiration. Plants and creatures present off CO2 while living and respiring and during lifeless and rotting. Plants are significant carbon sinks, gaining up enormous volumes of CO2 through the manner of photosynthesis.
While plants also discharge CO2 during the means of respiration, the volume of CO2 exercised up by plants by photosynthesis and discharged by exhalation approximately matches out. Volcanic action, forest wildfires, and diverse anthropological exercises deliver carbon.
In accordance with Dalton's Law of multiple proportions
<h3>Further explanation</h3>
Given
6.0g of carbon
22.0g or 14.0g of product
Required
related laws
Solution
the amount of air present ⇒ as an excess or limiting reactant
- air(O₂) as a limiting reactant(product=14 g)
C+0.5O₂⇒CO
6 + 8 = 14 g
mol O₂=8 g : 32 g/mol=0.25
mol C = 6 g : 12 g/mol = 0.5(2 x mol O₂)
mol CO= 2 x mol O₂ = 0.5 mol = 0.5 x 28 g/mol = 14 g
- air(O₂) as an excess reactant(product=22 g) an C as a limiting reactant
C+O₂⇒CO₂
6 + 16 = 22 g
mol C = 6 g : 12 g/mol = 0.5
mol O₂ = 16 g : 32 g/mol=0.5
mol CO₂ = 22 g : 44 g/mol = 0.5
if the mass firs element (C) constant, then the mass of the second element(O) in the two compounds will have a ratio as a simple integer.
CO = 6 : 8
CO₂ = 6 : 16
the ratio O = 8 : 16 = 1 : 2
In accordance with Dalton's Law of multiple proportions
