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2 years ago

Is platinum reactive or I reactive element and why ?

Chemistry

1 answer:

andre [41]2 years ago

6 0

Answer: It is reactive

Explanation:

Platinum is the least reactive element on the reactive scale.

For example, the platinum doesn't give the reaction with the oxygen that is found in the air, even though if it is heated but platinum does show the reaction with fluorine, chlorine and at red heat. Platinum can be found in a pure form in nature.

Platinum is reactive, it can react with some elements, but it is the least reactive element and also it is called unreactive, it doesn't easily react with elements.

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How many atoms it would take to make up the thickness of a banknote.

natita [175]

Answer:

Since one banknote has a thickness of about 0.1 millimeters, if all the banknotes produced in a year were stacked up, they would reach a height of approximately 300 kilometers, which is about 80 times the height of Mount Fuji.

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3 years ago

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2 years ago

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What happened to the reaction below

Citrus2011 [14]

c the temperature is increased?5)

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3 years ago

Given the following thermodynamic data, calculate the lattice energy of LiCl:

tiny-mole [99]

Answer:

$\boxed{\text{-862 kJ/mol}}$

Explanation:

One way to calculate the lattice energy is to use Hess's Law.

The lattice energy U is the energy released when the gaseous ions combine to form a solid ionic crystal:

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s); U = ?

We must generate this reaction rom the equations given.

(1) Li(s) + ½Cl₂ (g) ⟶ LiCl(s); ΔHf° = -409 kJ·mol⁻¹

(2) Li(s) ⟶ Li(g); ΔHsub = 161 kJ·mol⁻¹

(3) Cl₂(g) ⟶ 2Cl(g) BE = 243 kJ·mol⁻¹

(4) Li(g) ⟶Li⁺(g) +e⁻ IE₁ = 520 kJ·mol⁻¹

(5) Cl(g) + e⁻ ⟶ Cl⁻(g) EA₁ = -349 kJ·mol⁻¹

Now, we put these equations together to get the lattice energy.

(5) Li⁺(g) +e⁻ ⟶ Li(g) 520

(6) Li(g) ⟶ Li(s) -161

(7) Li(s) + ½Cl₂(g) ⟶ LiCl(s) -409

(8) Cl(g) ⟶ ½Cl₂(g) -121.5

(9) Cl⁻(g) ⟶ Cl(g) + e⁻ <u>+349</u>

Li⁺(g) + Cl⁻(g) ⟶ LiCl(s) -862

The lattice energy of LiCl is $\boxed{\textbf{-862 kJ/mol}}$ .

3 0

2 years ago

Read the following chemical equations.

lora16 [44]

This problem is providing two reduction-oxidation (redox) reactions in which the oxidized and reduced species can be identified by firstly setting the oxidation number of each element:

Reaction 1: 2K⁺I⁻ + H₂⁺O₂⁻ ⇒2K⁺O⁻²H⁺ + I₂⁰

Reaction 2: Cl₂⁰ + H₂⁰ ⇒ 2H⁺CI⁻

Next, we can see that iodine is being oxidized and oxygen reduced in reaction #1 and chlorine is being reduced and hydrogen oxidized in reaction #2 because the oxidized species increase the oxidation number whereas the reduced ones decrease it.

In such a way, the correct choice is C.

Learn more:

(Redox reactions) brainly.com/question/13978139

8 0

2 years ago