Answer:
<u><em></em></u>
- <u><em>pOH = 0.36</em></u>
Explanation:
Both <em>potassium hydroxide</em> and <em>lithium hydroxide </em>solutions are strong bases, so you assume 100% dissociation.
<u>1. Potassium hydroxide solution, KOH</u>
- Volume, V = 304 mL = 0.304 liter
- number of moles, n = M × V = 0.36M × 0.304 liter = 0.10944 mol
- 1 mole of KOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.10944
<u>2. LIthium hydroxide, LiOH</u>
- Volume, V = 341 mL = 0.341 liter
- number of moles, n = M × V = 0.341 liter × 0.51 M = 0.17391 mol
- 1mole of LiOH produces 1 mol of OH⁻ ion, thus the number of moles of OH⁻ is 0.17391
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<u>3. Resulting solution</u>
- Number of moles of OH⁻ ions = 0.10944 mol + 0.17391 mol = 0.28335 mol
- Volume of solution = 0.304 liter + 0.341 liter = 0.645 liter
- Molar concentration = 0.28335 mol / 0.645 liter = 0.4393 M
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<u>4. </u><em><u>pOH</u></em>
← answer
Kinetic energy due to the movement
Answer:
0.501 M
Explanation:
First we <u>convert 1.513 g of KHP into moles</u>, using its <em>molar mass</em>:
- 1.513 g ÷ 204.22 g/mol = 7.41x10⁻³ mol = 7.41 mmol
As <em>1 mol of KHP reacts with 1 mol of NaOH</em>, in 14.8 mL of the NaOH solution there were 7.41 mmoles of NaOH.
With the above information in mind we can <u>calculate the molarity of the NaOH solution</u>:
- 7.41 mmol / 14.8 mL = 0.501 M
Answer:
146 g/mol → option b.
Explanation:
This is a problem about the freezing point depression. The formula for this colligative property is:
ΔT = Kf . m . i
We assume i = 1, so our compound is not electrolytic.
ΔT = Freezing T° of pure solvent - Freezing T° of solution = 1.02 °C
m = molality (mol of solute/kg of solvent)
We convert the grams of solvent (benzene) to kg → 250 g . 1 kg/1000 = 0.250 kg.
We replace → 1.02°C = 5.12°C/mol/kg . mol/ 0.250kg . 1
1.02°C / 5.12 mol/kg/°C = mol/ 0.250kg
0.19922 mol/kg = mol/ 0.250kg
mol = 0.19922 . 0.250kg → 0.0498 mol
molar mass = g/mol → 7.27 g / 0.0498mol = 146 g/mol
Answer:
Expose the paper to the atmosphere or moisture
Explanation:
An interesting experiment that demonstrates the equilibrium of complexes is that of the invisible ink.
The invisible ink is pink colored [Co(H2O)6]Cl2 which is essentially colorless and pale when it is used to write on paper. This complex is almost colorless when dilute. Therefore, when it is used in writing, the writing can not be seen.
However, if the paper is left to stand or allowed to absorb moisture; the following equilibrium is set up:
[Co(H2O)6]Cl2 ⇄ [CoCl2(H2O)2] + 4H2O
The formation of blue [CoCl2(H2O)2] on standing or exposure to moisture enables the colored writing to be easily read.