Answer:
Order of decreasing: CaS> LiCl> CsCl. That is CsCl has the lowest lattice energy.
Explanation:
LATTICE ENERGYcan be used to estimate the STRENGHT of the bonds in an ionic compound.
ATOMIC RADIUS is a function of lattice energy. The atomic radius INCREASES as you move DOWN a group. LATTICE ENERGY DECREASES as ATOMIC RADIUS increases.
Considering the cations Lithium +1, and Caesium +1 , as one move DOWN the GROUP the ions get larger, this causes the LATTICE ENERGY TO DECREASE DOWN THE GROUP. This means that between lithium in and caesium ion, the Caesium ion has LOWER LATTICE ENERGY as COMPARE TO LITHIUM ION.
AS ONE MOVE ACROSS THE PERIOD, POSITIVE IONS BECOMES MORE CHARGED, and the MORE THE CHARGE, THE GREATER THE LATTICE ENERGY.
Therefore, Calcium ion will have higher lattice energy than Lithium ion.
Answer:
The name of it is Sulfur Hexafluoride
Explanation:
Bonus fact: It is over 5 times heavier than air so when inhaling it, it makes your voice go very deep. The opposite of helium some would say
Answer:
oh I'm here thanks for your points
Answer:
5×10⁵ L of ammonia (NH3)
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
N2 + 3H2 —> 2NH3
From the balanced equation above, we can say that:
3 L of H2 reacted to produce 2 L of NH3.
Finally, we shall determine the volume of ammonia (NH3) produced by the reaction of 7.5×10⁵ L of H2. This can be obtained as illustrated below:
From the balanced equation above,
3 L of H2 reacted to produce 2 L of NH3.
Therefore, 7.5×10⁵ L of H2 will react to produce = (7.5×10⁵ × 2)/3 = 5×10⁵ L of NH3.
Thus, 5×10⁵ L of ammonia (NH3) is produced from the reaction.
The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).
The general reaction is:
2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) (1)
We can write the above reaction in <u>two reactions</u>, one for oxidation and the other for reduction:
Li⁰(s) → Li⁺(aq) + e⁻ (2)
Fe²⁺(aq) + 2e⁻ → Fe⁰(s) (3)
We can see that Li⁰ is oxidizing to Li⁺ (by <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).
We must remember that the reducing agent is the one that will be oxidized by <u>reducing another element</u> and that the oxidizing agent is the one that will be reduced by <u>oxidizing another species</u>.
In reaction (1), the<em> reducing agent</em> is <em>Li</em> (it is oxidizing to Li⁺), and the <em>oxidizing agent </em>is<em> Fe(CH₃COO)₂</em> (it is reducing to Fe⁰).
Therefore, the reducing agent in reaction (1) is lithium (Li).
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