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qwelly [4]
3 years ago
10

The freons are a class of compounds containing carbon, chlorine, and fluorine. while they have many valuable uses, they have bee

n shown to be responsible for depletion of the ozone in the upper atmosphere. in 1991, two replacement compounds for freons went into production: hfc-134a () and hcfc-124 (). calculate the molar masses of these two compounds.
Chemistry
1 answer:
laila [671]3 years ago
3 0

1) HFC-134a is 1,1,1,2-tetrafluoroethane, with molecular formula C₂H₂F₄.

M(C₂H₂F₄) = 2Ar(C) + 2Ar(H) + 4Ar(F) · g/mol.

M(C₂H₂F₄) = 2 · 12 + 2 · 1.01 + 4 · 19 · g/mol.

M(C₂H₂F₄) = 102.02 g/mol; molar mass of HFC-134a.

2) 1) HCFC-124 is 1-chloro-1,2,2,2-tetrafluoroethane, with molecular formula C₂HClF₄.

M(C₂HClF₄) = 2Ar(C) + Ar(H) + Ar(Cl) + 4Ar(F) · g/mol.

M(C₂HClF₄) = 2 · 12 + 1.01 + 35.45 + 4 · 19 · g/mol.

M(C₂HClF₄) = 136.46 g/mol; molar mass of HCFC-124.

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Determine the pH of the resulting solution if 25 mL of 0.400 M strychnine (C21H22N2O2) is added to 50 mL of 0.200 M HCl? Assume
DIA [1.3K]

Answer:

pH = 4.56

Explanation:

The strychnine reacts with HCl as follows:

C₂₁H₂₂N₂O₂ + HCl ⇄ C₂₁H₂₂N₂O₂H⁺ + Cl⁻

<em />

For strychnine buffer:

pOH = 5.74 + log [C₂₁H₂₂N₂O₂H⁺] / [C₂₁H₂₂N₂O₂]

Initial moles of C₂₁H₂₂N₂O₂ are:

0.025L * (0.400 mol / L) = 0.01 moles C₂₁H₂₂N₂O₂

And of HCl are:

0.05L * (0.200 mol / L) = 0.01 moles HCl

That means after the reaction, you will have just 0.01 moles of C₂₁H₂₂N₂O₂H⁺ in 50mL + 25mL = 0.075L. And molarity is:

[C₂₁H₂₂N₂O₂H⁺] = 0.01 mol / 0.075L = 0.1333M

This conjugate acid, is in equilibrium with water as follows:

C₂₁H₂₂N₂O₂H⁺(aq) + H₂O(l) ⇄ C₂₁H₂₂N₂O₂ + H₃O⁺

<em />

<em>Where Ka = Kw / Kb = 1x10⁻¹⁴ / 1.8x10⁻⁶ = 5.556x10⁻⁹</em>

<em />

Ka is defined as:

Ka = 5.556x10⁻⁹ = [C₂₁H₂₂N₂O₂] [H₃O⁺] / [C₂₁H₂₂N₂O₂H⁺]

In equilibrium, concentrations are:

C₂₁H₂₂N₂O₂ = X

H₃O⁺ = X

C₂₁H₂₂N₂O₂H⁺ = 0.1333M - X

Replacing in Ka expression:

5.556x10⁻⁹ = [X] [X] / [0.1333M - X]

7.39x10⁻¹⁰ - 5.556x10⁻⁹X = X²

7.39x10⁻¹⁰ - 5.556x10⁻⁹X - X² = 0

Solving for X:

X = - 2.72x10⁻⁵M → False solution. There is no negative concentrations

X = 2.72x10⁻⁵M → Right solution.

As H₃O⁺ = X

H₃O⁺ = 2.72x10⁻⁵M

And pH = -log H₃O⁺

<h3>pH = 4.56</h3>
4 0
3 years ago
How many protons does Royalty, R-13, have?
dezoksy [38]
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4 0
3 years ago
When 15.3 g of sodium nitrate, NaNO3,was dissolved in water in a calorimeter, the temperature fell from 25.00oC to 21.56oC. If t
satela [25.4K]

Answer:

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

Explanation:

Heat lost by solution ad calorimeter = Q

Heat capacity of solution ad calorimeter = C = 1071 J/°C

Change in temperature = ΔT = 21.56°C - 25.00°C = -3.44°C

Q=C\times Delta T

Q=1071 J/^oC\times (-3.44^oC)=-3,684.24 J

Heat gained by sodium nitrate = -Q = -(-3,684.24 J)=3,684.24 J

Moles of sodium nitrate = \frac{15.3 g}{85 g/mol}=0.18 mol

When 0.18 mole of sodium nitrate was dissolved in water 3,684.24 joulesof heat was absorbed by it.

Then heat absorbed by 1 mole of sodium nitrate :

\frac{3,684.24 J}{0.18}=20,468 J=20.468 kJ

1 J = 0.001 kJ

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

8 0
3 years ago
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Basile [38]

Answer:

An element that is oxidized is a reducing agent, because the element loses electrons, and an element that is reduced is an oxidizing agent, because the element gains electrons.

7 0
3 years ago
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Carbon burns in the presence of oxygen to give carbon dioxide. Which chemical equation describes this reaction?
sasho [114]
2.) carbon + oxygen → carbon dioxide 
6 0
4 years ago
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