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
Rate = k * [A]^2 * [B]^1
<span>Use the data from any trial to calculate k. </span>
<span>k = (rate)/([A]^2 * [B]^1) </span>
<span>E.g., for Trial 1, we have </span>
<span>rate = 3.0×10−3 M/s </span>
<span>[A] = 0.50 M </span>
<span>[B] = 0.010 M </span>
<span>Plug those numbers in and crank out the answer. </span>
<span>Now with the calculated value of k, calculate the initial rate for [A] = 0.50 M and [B] = 0.075 M </span>
<span>rate = k * [A]^2 * [B]^1 </span>
<span>k = calculated value </span>
<span>[A] = 0.50 M </span>
<span>[B] = 0.075 M</span>
Packing
Explanation:
Ionic crystals are brittle due to the tight packing of their crystals. This provides little to no mobility between one another.
- Brittleness implies having little to no elasticity.
- Ionic crystals are held together by strong crystal lattice forces
- These forces prevents crystals from freely rotating and moving space.
- It holds them rigidly and fixed in place.
- Ionic crystals are hard and held electrostatic forces.
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Ionic compounds brainly.com/question/6071838
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Answer is: an oxybromate compound is KBrO₄ (x = 4).
ω(Br) = 43.66% ÷ 100%.
ω(Br) = 0.4366; mass percentage of bromine.
If we take 100 grams of compound:
m(Br) = ω(Br) · 100 g.
m(Br) = 0.4366 · 100 g.
m(Br) = 43.66 g; mass of bromine.
n(Br) = m(Br) ÷ M(Br).
n(Br) = 43.66 g ÷ 79.9 g/mol,
n(Br) = 0.55 mol; amoun of bromine.
From chemical formula (KBrOₓ), amount of potassium is equal to amount of bromine: n(Br) = n(K).
m(K) = 0.55 mol · 39.1 g/mol.
m(K) = 21.365 g; mass of potassium in the compound.
m(O) = 100 g - 21.365 g - 43.66 g.
m(O) =34.97 g; mass of oxygen.
n(O) = 34.97 g ÷ 16 g/mol.
n(O) = 2.185 mol.
n(K) : n(Br) : n(O) = 0.55 mol : 0.55 mol : 2.185 mol /÷ 0.55 mol.
n(K) : n(Br) : n(O) = 1 : 1 : 4.
Example of solid - solid homogeneous mixture is copper metal - silver metal like coins and alloys.
Homogeneous mixture is a mixture in which one of the substances often changes in form as in a solution of sugar in water. It contains variable proportions. Solution can contain two substances, three substances or more, in a single physical state. The component of a solution that is present in greatest quantity is usually called the solvent and all other components are called solutes.