D. 0.2 M
The concentration of a solution is basically the ratio of the solute present to the solvent in the solution. This is an intrinsic property, independent of the amount of solution that is present. A similar example is that of density. No matter the size of a sample, the density and concentration of that sample remain constant.
the primary consumer in those photos would be C
Answer: 4.
and 
Explanation:
a) The given reaction is 
As the mass on both reactant and product side must be equal:


As the atomic number on both reactant and product side must be equal:



b) 
Total mass on reactant side = total mass on product side
15 =15 + x
x = 0
Total atomic number on reactant side = total atomic number on product side
8 = 7 + y
y = 1

Mg reaction with O₂ gas will produce MgO so the equation will be
2Mg+O₂⇒2MgO. (You have to find the equation in order two figure out the number of moles of O₂ that will react with 1 mole of MgO).
The first step is to find the number of moles of Mg in 4.03g of Mg. You can do this by dividing 4.03g Mg by its molar mass (which is 24.3g/mol) to get 0.1658mol Mg. Then you have to find the number of moles of O₂ that will react with 0.1658mol Mg. To do this you need to use the fact that 1mol O₂ will react with 2mol Mg (this reatio is from the chemical equation) so you have to multiply 0.1658mol Mg by (1mol O₂)/(2mol Mg) to get 0.0829mol O₂. From here you would usually use PV=nRT and solve for V However, the question tells us that we are at STP, that means you can use the fact that 22.4L of gas is 1 mol of gas at STP. Using that information we can find the volume of O₂ gas by mulitlying 0.0829mol O₂ by 22.4L/mol to get 1.857L which equals 1857mL.
therefore, 1857mL of O₂ gas will react with 4.03g of Mg.
I hope this helps. Let me know in the comments if anything is unclear.
D. They all contain carbon as an important part of their structure.