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This doesn't need an ICE chart. Both will fully dissociate in water.
Assume HClO4 and KOH reacts with one another. All you need to do is determine how much HClO4 will remain after the reaction. Calculate pH.
Step 1:
write out balanced equation for the reaction
HClO4+KOH ⇔ KClO4 + H2O
the ratio of HClO4 to KOH is going to be 1:1. Each mole of KOH we add will fully react with 1 mole of HClO4
Step 2:
Determining the number of moles present in HClO4 and KOH
Use the molar concentration and the volume for each:
25 mL of 0.723 M HClO4
Covert volume from mL into L:
25 mL * 1L/1000mL = 0.025 L
Remember:
M = moles/L so we have 0.025 L of 0.723 moles/L HClO4
Multiply the volume in L by the molar concentration to get:
0.025L x 0.723mol/L = 0.0181 moles HClO4.
Add 66.2 mL KOH with conc.=0.273M
66.2mL*1L/1000mL = .0662 L
.0662L x 0.273mol/L = 0.0181 moles KOH
Step 3:
Determine how much HClO4 remains after reacting with the KOH.
Since both reactants fully dissociate and are used in a 1:1 ratio, we just subtract the number of moles of KOH from the number of moles of HClO4:
moles HClO4 = 0.0181; moles KOH = 0.0181, so 0.0181-0.0181 = 0
This means all of the HClO4 is used up in the reaction.
If all of the acid is fully reacted with the base, the pH will be neutral = 7.
Determine the H3O+ concentration:
pH = -log[H3O+]; [H3O+] = 10-pH = 10-7
The correct answer is 1.0x10-7.
The concentration of the hydrogen ions from molarity can be given with the number of hydrogen atoms in the molecular formula.
<h3>What is molarity?</h3>
Molarity is given as the moles of the solute present in a liters of solution. The compound with the number of hydrogen atoms in the molecular formula with degradation possesses the equivalent concentration of the hydrogen ions.
The compound with molecular formula AH having molarity 2, will produce 2M of hydrogen ions.
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The ratio of effusion rates for the lightest gas H₂ to the heaviest known gas UF₆ is 13.21 to 1
<h3>What is effusion?</h3>
Effusion is a process by which a gas escapes from its container through a tiny hole into evacuated space.
Rate of effusion ∝ 1/√Ц, (where Ц is molar mass)
Rate H₂ = 1/√ЦH₂
Rate UF₆ = 1/√ЦUF₆
Therefore, Rate H₂/ Rate UF₆ = √ЦH₂/√ЦUF₆
ЦH₂= 2.016 g/mol
ЦUF₆= 352.04 g/mol
Rate H₂ / Rate UF₆ = √352.04/√2.016 = 18.76/1.42
Rate H₂ / Rate UF₆ = 13.21
Therefore, H₂ is lower mass than UF₆. Thus H₂ gas will effuse 13 times more faster than UF₆ because the most probable speed of H₂ molecule is higher; therefore, more molecules escapes per unit time.
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