Photosynthesis is the process by which CO₂ is converted to an organic compound. The complete reaction is as follows;
6CO₂ + 6H₂O --> C₆H₁₂O₆ + 6O₂
stoichiometry of CO₂ to C₆H₁₂O₆ is 6:1
For 1 mol of C₆H₁₂O₆ to be formed - 6 mol of CO₂ is required
Therefore for 3.21 mol of C₆H₁₂O₆ to be formed - 6 x 3.21 mol of CO₂ required
Number of moles required are - 19.26 mol of CO₂
Volume = nRT/P
n = number of particles (moles)
R = universal gas constant (0.0821)
T = temperature (Kelvin)
P = pressure (atm)
(Assuming you have 1 mole of Helium in a chemical reaction) We would need to convert grams to moles: 12.0g He x 1 mol He/4 molar mass of He = 3 mol He
Convert Celsius to Kelvin: 100*C + 273.15 = 373.15 K
Now we can set up the equation for volume: (3mol)(0.0821)(373.15)/1.2atm = 76.6 L of Helium gas
<em>c</em> = 1.14 mol/L; <em>b</em> = 1.03 mol/kg
<em>Molar concentration
</em>
Assume you have 1 L solution.
Mass of solution = 1000 mL solution × (1.19 g solution/1 mL solution)
= 1190 g solution
Mass of NaHCO3 = 1190 g solution × (7.06 g NaHCO3/100 g solution)
= 84.01 g NaHCO3
Moles NaHCO3 = 84.01 g NaHCO3 × (1 mol NaHCO3/74.01 g NaHCO3)
= 1.14 mol NaHCO3
<em>c</em> = 1.14 mol/1 L = 1.14 mol/L
<em>Molal concentration</em>
Mass of water = 1190 g – 84.01 g = 1106 g = 1.106 kg
<em>b</em> = 1.14 mol/1.106 kg = 1.03 mol/kg
The material that has been widely used in casting has been copper. Thus, option C is correct.
Casting has been the process of molding the liquid into a specific shape with the mold. The casting involves the heating of elements and converting them into liquid. The liquid has been poured into the mold and cooling results in the casted material.
The casted material has been selected based on the boiling temperature, cooling temperature, resistivity, cost-effectiveness, and damping ability.
The material that has been consisted of these characteristics has been copper. Thus, copper has been commonly used for casting. Thus, option C is correct.
For more information about casting, refer to the link:
brainly.com/question/1253405
