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Answer:
0.1 M
Explanation:
The following data were obtained from the question:
Initial volume (V1) = 100 mL
Initial concentration (C1) = 0.5 M
Final volume (V2) = 500 mL
Final concentration (C2) =?
Using the dilution formula C1V1 = C2V2, the new concentration of the solution can be obtained as follow:
C1V1 = C2V2
0.5 × 100 = C2 × 500
50 = C2 × 500
Divide both side by 500
C2 = 50/500
C2 = 0.1 M
Therefore, the new concentration of the solution is 0.1 M
<u>Given:</u>
Mass of calcium nitrate (Ca(NO3)2) = 96.1 g
<u>To determine:</u>
Theoretical yield of calcium phosphate, Ca3(PO4)2
<u>Explanation:</u>
Balanced Chemical reaction-
3Ca(NO3)2 + 2Na3PO4 → 6NaNO3 + Ca3(PO4)2
Based on the reaction stoichiometry:
3 moles of Ca(NO3)2 produces 1 mole of Ca3(PO4)2
Now,
Given mass of Ca(NO3)2 = 96.1 g
Molar mass of Ca(NO3)2 = 164 g/mol
# moles of ca(NO3)2 = 96.1/164 = 0.5859 moles
Therefore, # moles of Ca3(PO4)2 produced = 0.0589 * 1/3 = 0.0196 moles
Molar mass of Ca3(PO4)2 = 310 g/mol
Mass of Ca3(PO4)2 produced = 0.0196 * 310 = 6.076 g
Ans: Theoretical yield of Ca3(PO4)2 = 6.08 g
The answer from this questions is the letter
B. He should not use his opinions as evidence.
