Answer:
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- <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
Answer:
How does the equilibrium change with the removal of hydrogen (H2) gas from this equation? 2H2S ⇌ 2H2(g) + S2(g) A. ... Equilibrium shifts left to produce less reactant.
Explanation:
option A is the correct answer
Equilibrium shifts right to produce more product.
I hope it will help you.
The lewis structure is helpful in showing how the bonding between atoms of a molecule are. The lewis structure of ammonia would be that the nitrogen atom will share three pairs of electron with the three hydrogen atoms leaving nitrogen to have 1 lone pair.<span />
Answer:
Vapor pressure of the solution is 22.9 torr at 25°C
Explanation:
When a non volatile solute is added to a pure solvent, the vapour pressure of the mixture decrease regard to the pure solvent following the equation (Raoult's law):
Where vapour pressure of the pure solvent (Water) is 23.8torr at 25°C.
Mole fraction of water is the ratio between moles of water and total moles in solution. The solution has a mole fraction of water of:
Replacing in Raoult's law:
<h3>Vapor pressure of the solution is 22.9 torr at 25°C</h3>
<span>the balanced equation for the reaction is as follows ;
Na</span>₂S + 2AgNO₃ ---> 2NaNO₃ + Ag₂<span>S
stoichiometry of Na</span>₂S to AgNO₃<span> is 1:2
number of AgNO</span>₃<span> moles reacted - 0.315 mol/L x 0.04000L = 0.0126 mol according to molar ratio of 1:2
number of Na</span>₂S moles required are - 1/2 x number of AgNO3 moles reacted Na₂<span>S moles = 0.0126 mol /2 = 0.00630 mol
molarity of Na</span>₂<span>S - 0.260 M
there are 0.260 mol in 1 L
therefore 0.00630 mol are in - 0.00630 mol / 0.260 mol/L
volume of Na</span>₂<span>S required = 0.0242 L
volume of Na</span>₂S required = 24.4 mL