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tatiyna
3 years ago
6

Phosgene is a potent chemical warfare agent that is now outlawed by international agreement. It decomposes by the reaction: COCl

2 (g) ⇋ CO (g) + Cl2 (g) Kc = 7.5 x 10-5 (at 362°C) Calculate the concentration of CO when 7.73 mol of phosgene decomposes and reaches equilibrium in a 10.0 liter flask. Multiply your answer by 103 and enter that number to 2 decimal places. HINT: Look at sample problem 17.9 in the 8th ed of Silberberg. Write a Kc expression. Calculate the initial concentration of phosgene. Do an ICE chart. Find the concentration of CO.
Chemistry
1 answer:
Oksana_A [137]3 years ago
6 0

Answer:

[CO] = 7.61x10⁻³M

7.61x10⁻³x10³ = 7.61

Explanation:

For a generic equation aA + bB ⇄ cC + dD, the constant of equilibrium (Kc) is:

Kc = \frac{[C]^cx[D]^d}{[A]^ax[B]^b}

We need to know the molar concentrations in the equilibrium. In the beginning, there is only COCl₂, and its concentration is the number of moles divided by the volume:

[COCl₂] = 7.73/10.0 = 0.773 M

So, the equilibrium will be:

COCl₂(g) ⇆ CO(g) + Cl₂(g)

0.773             0           0      <em>Initial</em>

-x                    +x         +x     <em> Reacts</em>

0.773-x            x           x       <em>Equilibrium</em>

Supposing that x<<0.773, then:

Kc = \frac{x*x}{0.773}

7.5x10⁻⁵ = x²/0.773

x² = 5.7975x10⁻⁵

x = √5.7975x10⁻⁵

x = 7.61x10⁻³ M

The supposing is correct, so [CO] = 7.61x10⁻³ x 10³ = 7.61

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2 years ago
Calculate the MASS if the density is 2.34 g/mL and the volume is 12.50 mL.
AlexFokin [52]

Answer:

<h2>mass = 29.25 g</h2>

Explanation:

The denisty of a substance can be found by using the formula

Density =  \frac{mass}{volume}

Making mass the subject we have

<h3>mass = Density × volume</h3>

From the question

Density = 2.34 g/mL

volume = 12.50 mL

Substitute the values into the above formula and solve for the mass

That's

mass = 2.34 × 12.50

We have the final answer as

<h3>mass = 29.25 g</h3>

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3 0
3 years ago
Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium nitrate. A soluti
koban [17]

Answer:

1) 2.0 g

2) 0 g

3) 4.17 g

4) 2.57 g

Explanation:

First of all, we need to know the compounds and the reaction. The ion carbonate is CO3^{-2}, and the ion nitrate is NO3^{-}.

Sodium is in group 1, so it must lose one electron to be stable, and be the cation Na^{+}. Silver has only one electron too, so the cation will be Ag^{+}.

To form the chemical compounds, first we put the cation, then the anion, and change their charges without the signal:

Sodium carbonate: Na2CO3

Silver nitrate: AgNO3

Silver carbonate: Ag2CO3

Sodium nitrate: NaNO3

The balanced reaction will be:

Na2CO3 + 2 AgNO3 --> Ag2CO3 + 2 NaNO3

Now, we must check the stoichiometry, which will be 1:2:1:2 (always in number of moles)

The question wants to know the mass value, so we need to know the molar mass of these compounds. Checking the periodic table will see that:

Na = 23 g/mol, C = 12 g/mol, N = 14 g/mol, O = 16 g/mol, Ag = 108 g/mol

So the molar mass of the compounds must be:

Na2CO3 = 106 g/mol (2x23 + 12 + 3x16)

AgNO3 = 170 g/mol (108 + 14 + 3x16)

Ag2CO3 = 276 g/mol (2x108 + 12 + 3x16)

NaNO3 = 85 g/mol

We have a mixture of the reactants, so one probably would be in excess, so, first will need to test. Let's do the stoichiometry calculus using silver nitrate as the limit, so:

1 mol of Na2CO3 ---------- 2 mol of AgNO3

106 g ------------------------------ 2x170 = 340 g

x ------------------------------------ 5.14 g

By a simple direct three rule:

340x = 544.84

x = 1.6 g of Na2CO3

That means that for this reaction, we only need 1.6 g of Na2CO3 to react with 5.14 of AgNO3. How we have 3.60 g of Na2CO3, it is on excess, and all the AgNO3 will be consumed.

1) The mass of Na2CO3 that remains after the reaction will be the initial less the mass that reacted:

m = 3.6 - 1. 6 = 2.0 g

2) All the AgNO3 reacted, so there isn't a mass present after the reaction.

m = 0 g

3) Now, doing the stoichiometry calculus between AgNO3 and Ag2CO3

2 moles of AgNO3 ------------- 1 mol of Ag2CO3

2x170 g ------------------------------- 276 g

5.14 g --------------------------------- x

By a simple direct three rule:

340x = 1418.64

x = 4.17 g of Ag2CO3

4) Now, doing the stoichiometry calculus between AgNO3 and NaNO3

2 moles of AgNO3 ----------------------- 2 moles of NaNO3

2x170 g ---------------------------------------- 2x85 g

5.14 g ------------------------------------------- x

By a simple direct three rule:

340x = 873.8

x = 2.57 g

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V/T = ½V/x

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