I am guessing that your solutions of HCl and of NaOH have approximately the same concentrations. Then the equivalence point will occur at pH 7 near 25 mL NaOH.
The steps are already in the correct order.
1. Record the pH when you have added 0 mL of NaOH to your beaker containing 25 mL of HCl and 25 mL of deionized water.
2. Record the pH of your partially neutralized HCl solution when you have added 5.00 mL of NaOH from the buret.
3. Record the pH of your partially neutralized HCl solution when you have added 10.00 mL, 15.00 mL and 20.00 mL of NaOH.
4. Record the NaOH of your partially neutralized HCl solution when you have added 21.00 mL, 22.00 mL, 23.00 mL and 24.00 mL of NaOH.
5. Add NaOH one drop at a time until you reach a pH of 7.00, then record the volume of NaOH added from the buret ( at about 25 mL).
6. Record the pH of your basic HCl-NaOH solution when you have added 26.00 mL, 27.00 mL, 28.00 mL, 29.00 mL and 30.00 mL of NaOH.
7. Record the pH of your basic HCl-NaOH solution when you have added 35.00 mL, 40.00 mL, 45.00 mL and 50.00 mL of NaOH from your 50mL buret.
<span>Let's </span>assume that water vapor has ideal gas
behavior. <span>
Then we can use ideal gas formula,
PV = nRT<span>
</span><span>Where, P is the pressure of the gas (Pa), V
is the volume of the gas (m³), n is the number
of moles of gas (mol), R is the universal gas constant ( 8.314 J mol</span></span>⁻¹ K⁻¹) and T is temperature in Kelvin.<span>
<span>
</span>P = 1 atm = 101325 Pa (standard pressure)
V = 13.97 L = 13.97 x 10</span>⁻³ m³<span>
n = ?
R = 8.314 J mol</span>⁻¹ K⁻¹<span>
T = 0 °C = 273 K (standard temperature)
<span>
By substitution,
</span>101325 Pa x 13.97x 10</span>⁻³
m³ = n x 8.314 J mol⁻¹ K⁻¹ x 273 K<span>
n = 0.624 mol
<span>
Hence, the moles of water vapor at STP is 0.624 mol.
According to the </span></span>Avogadro's constant, 1 mole of substance has 6.022 × 10²³ particles.
<span>
Hence, number of atoms in water vapor = 0.624 mol x </span>6.022 × 10²³ mol⁻¹
<span> = 3.758 x 10</span>²³<span>
</span>
3 subatomic particles in the nucleus
which is proton(24),neutron (28) and electrons(24)
Answer:
27.6mL of LiOH 0.250M
Explanation:
The reaction of lithium hydroxide (LiOH) with chlorous acid (HClO₂) is:
LiOH + HClO₂ → LiClO₂ + H₂O
<em>That means, 1 mole of hydroxide reacts per mole of acid</em>
Moles of 20.0 mL = 0.0200L of 0.345M chlorous acid are:
0.0200L ₓ (0.345mol / L) = <em>6.90x10⁻³ moles of HClO₂</em>
To neutralize this acid, you need to add the same number of moles of LiOH, that is 6.90x10⁻³ moles. As the LiOH contains 0.250 moles / L:
6.90x10⁻³ moles ₓ (1L / 0.250mol) = 0.0276L of LiOH =
<h3>27.6mL of LiOH 0.250M</h3>
