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liq [111]
3 years ago
11

PLEASE HELP ASAP

Chemistry
1 answer:
Lapatulllka [165]3 years ago
8 0

Answer:

See explanation

Explanation:

We must first write the equation of the reaction as follows;

C3H8 + 5O2 ----> 3CO2 + 4H2O

Now;

We obtain the number of moles of C3H8 = 132.33g/44g/mol = 3 moles

So;

1 mole of C3H8 yields 3 moles of CO2

3 moles of C3H8 yields 3 × 3/1 = 9 moles of CO2

We obtain the number of moles of oxygen = 384.00 g/32 g/mol = 12 moles

So;

5 moles of oxygen yields 3 moles of CO2

12 moles of oxygen yields 12 × 3/5 = 7.2 moles of CO2

We can now decide on the limiting reactant to be C3H8

Therefore;

Mass of CO2 produced = 9 moles of CO2 × 44 g/mol = 396 g of CO2

Again;

1 moles of C3H8 yields 4 moles of water

3 moles of C3H8 yields 3 × 4/1 = 12 moles of water

Hence;

Mass of water = 12 moles of water × 18 g/mol = 216 g of water

In order to obtain the percentage yield from the reaction, we have;

b) Actual yield = 269.34 g

Theoretical yield = 396 g

Therefore;

% yield = actual yield/theoretical yield × 100/1

Substituting values

% yield = 269.34 g /396 g × 100

% yield = 68%

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Help Pls!
Salsk061 [2.6K]

Answer:

d= 50.23 g/cm³

Explanation:

Given data:

radius = 137.9 pm

mass is = 5.5 × 10−22 g

density = ?

Solution:

volume of sphere= 4/3π r³

First of all we calculate the volume:

v= 4/3π r3

v= 1.33× 3.14× (137.9)³

v= 1.33 × 3.14 × 2622362.939 pm³

v= 1.095 × 10∧7 pm³

v= 1.095 × 10∧-23 cm³

Formula:

Density:

d=m/v

d= 5.5 × 10−22 g/ 1.095 × 10∧-23 cm³

d= 5.023 × 10∧+1 g/cm³

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8 0
3 years ago
Based on the reactivities of the elements involved, which reactions will form products that are more stable than the reactants?
myrzilka [38]

2 LiI + Cl₂ → 2 LiCl + I₂

2 LiBr + F₂ → 2 LiF + Br₂

<h3>Explanation</h3>

Each of the five reactions involve one halogen molecule (F₂, Cl₂, Br₂, and I₂) substituting the ion of another halogen (F⁻, Cl⁻, Br⁻, and I⁻).

Halogen atoms are found in group 17 of the periodic table. They are all non-metal elements. Each of the halogen atom will gain one electron to form an ion of charge -1. However, the tendency to do so decreases down the group.

  • F is the first halogen in group 17. It has only two shells of electrons.
  • Cl is right under F. Its electrons occupy three main energy shells.
  • Br follows with four main energy shells.
  • I is under Br and has five main energy shells.

Atoms of all four elements have the same effective nuclear charge of +7. However, F has the smallest radius. As a result, it has the strongest hold on electrons around it. Its ion F⁻ is more stable than ions of Cl, Br, or I. Similarly, its molecule F₂ is more reactive than Cl₂, Br₂, and I₂.

As a result, the stability of halogen molecules increases down the group:

  • Stability: F₂ < Cl₂ < Br₂ < I₂.

The stability of halogen ions decreases down the group:

  • Stability: F⁻ > Cl⁻ > Br⁻ > I⁻.

Cl₂ repaces F⁻ (from LiF) in first reaction. F₂ and Cl⁻ are produced. F₂ is less stable than Cl₂. Cl⁻ is less stable than F⁻.

Cl₂ replaces I⁻ (from LiI) in the second reaction. I₂ and Cl⁻ are produced. I₂ is more stable than Cl₂. Cl⁻ is more stable than I⁻.

Br₂ replaces Cl⁻ (from LiCl) in the third reaction. Cl₂ and Br⁻ are produced. Cl₂ is less stable than Br₂. Br⁻ is less stable than Cl⁻.

F₂ replaces Br⁻ (from LiBr) in the fourth reaction. Br₂ and F⁻ are produced. Br₂ is more stable than F₂. F⁻ is more stable than Br⁻.

I₂ replaces Br⁻ (from LiBr) in the fifth reaction. Br₂ and I⁻ are produced. Br₂ is less stable than I₂. I⁻ is less stable than Br⁻.

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