5.5484 x 10^4
^= the tiny 4 at the top of the number
couldn't figure out how to make it so gave up
Answer:
0.1 mol
Explanation:
Step 1: Given data
- Volume of the solution of sodium hydroxide (V): 200 cm³
- Molar concentration of sodium hydroxide (C): 0.5 mol/dm³
Step 2: Convert "V" to dm³
We will use the conversion factor 1 dm³ = 1000 cm³.
200 cm³ × 1 dm³/1000 cm³ = 0.200 dm³
Step 3: Calculate the moles (n) of NaOH
The molarity of the NaOH solution is equal to the moles of NaOH divided by the volume of solution.
C = n/V
n = C × V
n = 0.5 mol/dm³ × 0.200 dm³ = 0.1 mol
The 2, written as a subscript, counts for only the hydrogen. 3 H2O has 6 atoms of H and 3 atoms<span> of O. The 3, written as a coefficient, means that there are 3 “H2O”s.</span>
<span>Mg(OH)2 has </span>1 atom<span> of Mg, 2 atoms of O, and 2 atoms of H.</span>
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:
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