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3 years ago

36. Write the SHORTCUT electron configuration for Caesium (Cs) which is element #55.

Chemistry

1 answer:

miv72 [106K]3 years ago

6 0

Answer:

Cesium. Full electron configuration of cesium: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s1. xenon ← cesium → barium

Explanation:

most include 36. :-]

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why can an increase in temperature lead to more effective collisions between reactant particles and an increase in the rate of a

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3 years ago

04.05 mol

victus00 [196]

Answer:

34.8 g

Explanation:

Answer:

We have the masses of two reactants, so this is a limiting reactant problem.

We will need a balanced equation with masses, moles, and molar masses of the compounds involved.

1. Gather all the information in one place with molar masses above the formulas and masses below them.

Mᵣ: 123.90 70.91 208.24

P₄ + 20Cl₂ ⟶ 4PCl₅

Mass/g: 46.0 32.0

2. Calculate the moles of each reactant

$\text{moles of P}_{4} = \text{46.0 g P}_{4} \times \dfrac{\text{1 mol P}_{4}}{\text{123.90 g P}_{4}} = \text{0.3713 mol P}_{4}\\\\\text{moles of Cl}_{2} = \text{32.0 g Cl }_{2} \times \dfrac{\text{1 mol Cl }_{2}}{\text{70.91 g Cl }_{2}} = \text{0.4513 mol Cl }_{2}$

3. Calculate the moles of PCl₅ we can obtain from each reactant

From P₄:

The molar ratio is 4 mol PCl₅:4 mol P₄

$\text{Moles of PCl}_{5} = \text{0.3713 mol P}_{4} \times \dfrac{\text{4 mol PCl}_{5}}{\text{4 mol P}_{4}} = \text{0.3713 mol PCl}_{5}$

From Cl₂:

The molar ratio is 4 mol PCl₅:20 mol Cl₂

$\text{Moles of PCl}_{5} = \text{0.4513 mol Cl}_{2}\times \dfrac{\text{4 mol PCl}_{5}}{\text{20 mol Cl}_{2}} = \text{0.090 26 mol PCl}_{5}$

4. Identify the limiting and excess reactants

The limiting reactant is chlorine, because it gives the smaller amount of PCl₅.

The excess reactant is phosphorus.

5. Mass of excess reactant

(a) Moles of P₄ reacted

The molar ratio is 1 mol P₄:20 mol Cl₂

$\text{Moles reacted} = \text{0.4513 mol Cl}_{2} \times \dfrac{\text{4 mol P}_{4}}{\text{20 mol Cl}_{2}} = \text{0.090 26 mol P}_{4}$

(b) Mass of P₄ reacted

$\text{Mass reacted} = \text{0.090 26 mol P}_{4} \times \dfrac{\text{123.90 g P}_{4}}{\text{1 mol P}_{4}} = \text{11.18 g P}_{4}$

(c) Mass of P₄ remaining

Mass remaining = original mass – mass reacted = (46.0 - 11.18) g = 34.8 g P₄

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4 years ago

Jonathan conducts an experiment to determine what solutes readily dissolve in water. He places 3 tablespoons of potting soil int

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Answer:

A. Mixture

Explanation:

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