Answer:
17
Explanation:
Step 1: Calculate the needed concentrations
[A]i = 1.00 mol/5.00 L = 0.200 M
[B]i = 1.80 mol/5.00 L = 0.360 M
[B]e = 1.00 mol/5.00 L = 0.200 M
Step 2: Make an ICE chart
A(aq) + 2 B(aq) ⇄ C(aq)
I 0.200 0.360 0
C -x -2x +x
E 0.200-x 0.360-2x x
Then,
[B]e = 0.360-2x = 0.200
x = 0.0800
The concentrations at equilibrium are:
[A]e = 0.200-0.0800 = 0.120 M
[B]e = 0.200 M
[C]e = 0.0800 M
Step 3: Calculate the concentration equilibrium constant (K)
K = [C] / [A] × [B]²
K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17
stirring, temperature, and particle size
Answer is: theoretical molarity of water is 55.1222 mol/L.<span>
d(H</span>₂O) = 0.9922 g/mL.
M(H₂O) = 2 · Ar(H) + Ar(O) · g/mol.
M(H₂O) = 2 + 16 · g/mol = 18 g/mol.
c(H₂O) = d(H₂O) ÷ M(H₂O).
c(H₂O) = 0.9922 g/mL ÷ 18 g/mol.
c(H₂O) = 0.0551 mol/mL.
c(H₂O) = 0.0551 mol/mL · 1000 mL/L = 55.1222 mol/L.
Answer: -
0.1 ml of bleach should be added to each liter of test solution.
Explanation:-
Let the volume of bleach to be added is B ml.
Density of stock solution = 1.0 g/ml
Mass of stock solution = Volume of stock x density of stock
= B ml x 1.0 g/ml
= B g
Amount of NaOCl in this stock solution = 5% of B g
=
x B g
= 0.05 B g
Now each test solution must be added 5 mg/l NaOCl.
Thus each liter of test solution must have 5 mg.
Thus 0.05 B g = 5 mg
= 0.005 g
B = 
= 0.1
Thus 0.1 ml of bleach should be added to each liter of test solution.