Answer:
0.172 M
Explanation:
The reaction for the first titration is:
First we <u>calculate how many HCl moles reacted</u>, using the <em>given concentration and volume</em>:
- 19.6 mL * 0.189 M = 3.704 mmol HCl
As one HCl mol reacts with one NaOH mol, <em>there are 3.704 NaOH mmoles in 25.0 mL of solution</em>. With that in mind we <u>determine the NaOH solution concentration</u>:
- 3.704 mmol / 25.0 mL = 0.148 M
As for the second titration:
- H₃PO₄ + 3NaOH → Na₃PO₄ + 3H₂O
We <u>determine how many NaOH moles reacted</u>:
- 34.9 mL * 0.148 M = 5.165 mmol NaOH
Then we <u>convert NaOH moles into H₃PO₄ moles</u>, using the <em>stoichiometric coefficients</em>:
- 5.165 mmol NaOH * = 1.722 mmol H₃PO₄
Finally we <u>determine the H₃PO₄ solution concentration</u>:
- 1.722 mmol / 10.0 mL = 0.172 M
Answer:
2.5, - 2.5
Explanation:
Given that :
Coordinate of A = 0
and AR = 5 ; distance between A and R
Possible. Coordinate of the midpoint of AR overbar.
Hence for a distance of 5 points between A and R, then R may be located 5 points either side of A, that is to the left of right.
Kindly check attached picture for detailed workings
Although it's true that MgO is only very slightly soluble in water, you can consider the mixture a conventional solution for the purposes of this question. The % mass is the mass of MgO divided by the mass of the entire solution:
45.8 / (45.8 + 205) * 100 = <span>18.3%</span>
O3 + M2+(aq) + H2O(l) => O2(g) + MO2(s) + 2 H+
Eo(cell) = Eo(O3/O2) - Eo(MO2/M2+)
0.44 = 2.07 - Eo(MO2/M2+)
Eo(MO2/M2+) = 1.59 V
Answer:
true
Explanation:
This sounds like the way an endothermic reaction is going to operate.
It's a very good description of taking on heat which is endothermic.
The statement is true.