Answer:
Explanation:
Flame test:
The metals ions can be detected through the flame test. Different ions gives different colors when heated on flame. Tom perform the flame test following steps should follow:
1. Dip a wire loop in the solution of compound which is going to be tested.
2. After dipping put the loop of wire on bunsen burner flame.
3. Observe the color of flame.
4. Record the flame color produce by compound
Color produce by metals:
Red = Lithium, zirconium, strontium, mercury, Rubidium (red violet)
Orange-red = calcium
Yellow = sodium, iron (brownish yellow)
Green = green
Blue = cesium. arsenic, copper, tantalum, indium, lead
Violet = potassium (lilac)
If I'm correct the answer should be a series circuit :) Hopefully this helps you out
Answer:
Francium (Fr)
Explanation:
From the given choices, francium will have the lowest ionization energy.
Ionization energy is the energy required to remove the most loosely held electron within an atom.
The magnitude of the ionization energy depends on the characteristics of the atom in relation to its nuclear charge, atomic radius, stability etc.
- Generally on the periodic table, ionization energy increases from left to right on the table
- As you go from metals to non-metals and to gases, the value of the ionization energy increases steadily.
- Down the group, the value reduces.
- Since Francium is the most metallic of all the given choices, it has the highest ionization energy.
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
