Answer:
a) The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) 0.0035 mole
c) 0.166 M
Explanation:
Phosphoric acid is tripotic because it has 3 acidic hydrogen atom surrounding it.
The equation of the reaction is expressed as:

1 mole 3 mole
The relationship at equivalence is that 1 mole of phosphoric acid will need three moles of sodium hydroxide.
b) if 10.00 mL of a phosphoric acid solution required the addition of 17.50 mL of a 0.200 M NaOH(aq) to reach the endpoint; Then the molarity of the solution is calculated as follows

10 ml 17.50 ml
(x) M 0.200 M
Molarity = 
= 0.0035 mole
c) What was the molar concentration of phosphoric acid in the original stock solution?
By stoichiometry, converting moles of NaOH to H₃PO₄; we have
= 
= 0.00166 mole of H₃PO₄
Using the molarity equation to determine the molar concentration of phosphoric acid in the original stock solution; we have:
Molar Concentration = 
Molar Concentration = 
Molar Concentration = 0.166 M
∴ the molar concentration of phosphoric acid in the original stock solution = 0.166 M
The correct answer is (D)
All the above are reasons the carbon , oxygen, and nitrogen cycles are vital to life on earth.
The explanation:
because :
1) Carbon, oxygen, and nitrogen are vital components of life on Earth.
2) The carbon, oxygen, and nitrogen cycles allow vital elements to return to usable form by organisms.
3)The carbon, oxygen, and nitrogen cycles are an important interface between biotic and abiotic factors
4) They are all biogeochemical cycles.
5) They all involve an interaction between living and nonliving elements.
6)They are all part of the Earth system.
Answer:false
Explanation:matter cannot be created or destroyed in a chemical reaction it is rearranged
C.) iron oxide <span>is responsible for the red color of Mars.
Hope this helps!</span>
Answer:
Part A: 36 MBq; Part B: 18 MBq
Explanation:
The half-life is the time it takes for half the substance to disappear.
The activity decreases by half every half-life
A =Ao(½)^n, where n is the number of half-lives.
Part A
3.0 da = 1 half-life
A = Ao(½) = ½ × 72 MBq = 36 MBq
Part B
6.0 da = 2 half-lives
A = Ao(½)^2 = ¼ × 72 MBq = 18 MBq