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2 years ago

A mixture of 0.600 mol of bromine and 1.600 mol of iodine is placed into a rigid 1.000-L container at 350°C.

Chemistry

1 answer:

nata0808 [166]2 years ago

4 0

The equilibrium constant for this reaction at 350°C is determined as 5.85.

<h3>Concentration of each component</h3>

concentration of bromine, C(Br) = 0.6 mol/1 = 0.6

concentration of iodine, C(I) = 1.6 mol/1 = 1.6

<h3>Create an ICE table</h3><h3>What is ICE table?</h3>

An ICE table is a tabular system of keeping track of changing concentrations in an equilibrium reaction.

ICE is an abbreviation that stands for initial, change, equilibrium.

Create ICE table for the reactants and products formed;

Br2(g) + I2(g) ↔ 2IBr(g)

I 0.6 1.6 0

C -1.19 -1.19 1.19

E 0.6 - 1.19 1.6 - 1.19 1.19

E = -0.59 0.41 1.19

<h3>Equilibrium constant </h3>

The equilibrium constant is calculated as follows;

KC = [IBr]²/[Br][I]

KC = (1.19²) / (0.59 x 0.41)

KC = 5.85

Thus, the equilibrium constant for this reaction at 350°C is determined as 5.85.

Learn more about equilibrium constant here: brainly.com/question/19340344

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A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

3 years ago

Need help fast!!

stepan [7]

Answer:

The answer is "5.18 ".

Explanation:

It's a question of chemistry. Therefore, the following solution is provided:

We will first determine the solution's pOH. The following can possible:

The concentration of Hydroxide ion $[OH^{-}] = 1.5\times 10^{-9}\ M$

$pOH =?\\\\pOH = -\log [OH^{-}]\\\\pOH = -\log 1.5\times 10^{-9}\\\\pOH = 8.82\\\\$

Furthermore, the pH of the solution shall be established. It was provided as follows:

$pOH = 8.82\\\\pH =?\\\\pH + pOH = 14\\\\pH + 8.82 = 14\\\\$

When collecting all the like terms:

$pH = 14 -8.82\\\\pH = 5.18$

Therefore, the solution of pH is 5.18.

8 0

3 years ago

3. Ethanol has a density of 0.800 g/mL. <br> a. What is the mass of 225 mL of ethanol?

dedylja [7]

Since the ethanol has a density of 0.800 g/mL, the mass of the ethanol is 180 grams.

<u>Given the following data:</u>

Volume of ethanol = 225 mL

Density of lead ball = 0.800 g/mL.

To find the mass of the ethanol;

Density can be defined as mass all over the volume of an object.

Mathematically, the density of a substance is given by the formula;

$Density = \frac{Mass}{Volume}$

Making mass the subject of formula, we have;

$Mass = Density \times Volume$

Substituting the given parameters into the formula, we have;

<em>Mass of ethanol </em><em>=</em><em> 180 grams.</em>

3 0

3 years ago

How many elements are found in 2CuSO4?<br> A.5<br> B.4<br> C.2<br> D.3

Aneli [31]

Answer:

Explanation:

the reason is you have Cu S and O

4 0

3 years ago

When 7.00 g of hydrogen react with 70.0 g of nitrogen, hydrogen is considered the limiting reactant because:

noname [10]

7.5 mol of hydrogen would be needed to consume the available nitrogen.

Explanation:

When hydrogen reacts with nitrogen, ammonia is formed as shown below;

3H₂ (g) + N₂ (g) → 2NH₃ (g)

As seen from the equation, every 3 moles of H₂ react with a mole of N₂ to form 2 moles of NH₃.

The limiting factor in a chemical reaction is the reactant that gets depleted first.

Because the molar mass of nitrogen gas is approximately 28g/mol, 70g of nitrogen gas would be 2.5 moles.

The reaction ratio of nitrogen to hydrogen in the reaction is 1 : 3. The reaction would require 2.5 * 3 (7.5) moles of hydrogen for a complete reaction.

However since there are only 7g on hydrogen, (Remember 1 mole of H₂ is approximately 2g), the available moles of H₂ is 7 / 2 = 3.5

3.5 moles fall short of the 7.5 moles of H₂ required for a complete reaction. H₂ gets depleted first before N₂. The reaction would require 4 more moles of H₂.

6 0

3 years ago