This is an acid base reaction and the chemical equation for the above reaction is as follows;
KOH + HClO₄ ---> KClO₄ + H₂O
the stoichiometry of acid to base is 1:1
KOH is a strong base and HClO₄ is a strong acid therefore they both ionize completely into their respective ions
Number of KOH moles - 0.723 M/1000 mL/L x 25.0 mL = 0.018 mol
Number of HClO₄ moles - 0.273 M/1000 mL/L x 50 mL = 0.013 mol
since acid and base react completely, 0.013 mol of acid reacts with 0.013 mol of base.
The excess base remaining is - 0.018 - 0.013 = 0.005 mol
total volume of solution = 25.0 mL + 50.0 mL = 75.0 mL
[OH⁻] = 0.005 mol/0.075 L = 0.067 M
pOH = -log[OH⁻]
pOH = -log(0.067 M)
pOH = 1.17
pOH + pH = 14
Therefore pH = 14 - 1.17 = 12.83
by knowing pH we can calculate the [H₃O⁺]
pH = -log [H₃O⁺]
[H₃O⁺] = antilog[-12.83]
[H₃O⁺]= 1.47 x 10⁻¹³ M
Answer:
Average atomic mass = 15.86 amu.
Explanation:
Given data:
Number of atoms of Z-16.000 amu = 205
Number of atoms of Z-14.000 amu = 15
Average atomic mass = ?
Solution:
Total number of atoms = 205 + 15 = 220
Percentage of Z-16.000 = 205/220 ×100 = 93.18%
Percentage of Z-14.000 = 15/220 ×100 = 6.82 %
Average atomic mass = (abundance of 1st isotope × its atomic mass) +(abundance of 2nd isotope × its atomic mass) / 100
Average atomic mass = (93.18×16.000)+(6.82×14.000) /100
Average atomic mass = 1490.88 + 95.48 / 100
Average atomic mass = 1586.36 / 100
Average atomic mass = 15.86 amu.
Answer:
The proportion of dissolved substances in seawater is usually expressed in ppm, ppb or ppt
Explanation:
The concentration of very diluted solutions should be expressed in parts per million, billion or trillion.
ppm = mass from the solute . 10⁶ / mass or volume of the solution
ppb = mass from the solute . 10⁹ / mass or volume of the solution
ppt = mass from the solute . 10¹² / mass or volume of the solution
ppm = mg/kg, μg/g, μg/mL → These are the units
ppb = ng/g
ppt = pg/g
Following reaction arise between Br2 and Cl2
Br2 + Cl2 → 2BrCl
(1mole) (1mole) (2moles)
From above balanced reaction, it can be seen that 1 mole of Br2 reacts with 1 mole of Cl2 to form 2 mole of BrCl
Thus, when <span>2.74 mol Cl2 reacts with excess Br2, 2.74 X 2 = 5.48 moles of BrCl will be formed. </span>
