Mendel crossed tall pea plants with short pea plants. the offspring were all tall plants.

Given the following balanced equation at 120°C: A(g) + B(g) ⇋ 2 C(g) + D(s)(a) At equilibrium a 4.0 liter container was found to

BlackZzzverrR [31]

Answer:

a) kc = 0,25

b) [A] = 0,41 M

c) [A] = 0,8 M

[B] =0,2 M

[C] = 0,2M

Explanation:

The equilibrium-constant expression is defined as the ratio of the concentration of products over concentration of reactants. Each concentration is raised to the power of their coefficient.

Also, pure solid and liquids are not included in the equilibrium-constant expression because they don't affect the concentration of chemicals in the equilibrium.

If global reaction is:

A(g) + B(g) ⇋ 2 C(g) + D(s)

The kc = $\frac{[C]^2}{[A][B]}$

a) The concentrations of each compound are:

[A] = $\frac{1,60 moles}{4,0 L}$ = 0,4 M

[B] = $\frac{0,40 moles}{4,0 L}$ = 0,1 M

[C] = $\frac{0,40 moles}{4,0 L}$ = 0,1 M

kc = $\frac{[0,1]^2}{[0,4][0,1]}$ = 0,25

b) The addition of B and D in the same amount will, in equilibrium, produce these changes:

[A] = $\frac{1,60-x moles}{4,0 L}$

[B] = $\frac{0,60-x moles}{4,0 L}$

[C] = $\frac{0,60+2x moles}{4,0 L}$

0,25 = $\frac{[0,60+2x]^2}{[1,60-x][0,60-x]}$

You will obtain

3,75x² +2,95x +0,12 = 0

Solving

x =-0,74363479081119 → No physical sense

x =-0,043031875855476

Thus, concentration of A is:

$\frac{1,60-(-0,04 moles)}{4,0 L}$ = 0,41 M

c) When volume is suddenly halved concentrations will be the concentrations in equilibrium over 2L:

[A] = $\frac{1,60 moles}{2,0 L}$ = 0,8 M

[B] = $\frac{0,40 moles}{2,0 L}$ = 0,2 M

[C] = $\frac{0,40 moles}{2,0 L}$ = 0,2M

I hope it helps!

8 0

4 years ago

Read 2 more answers

Need help ASAP!!!! what is the value (angle) for the C=C=O bond in Ketene i.e. CH2=C=O

stepan [7]

Answer:

$180^\circ$ by the VSEPR theory.

Explanation:

This question is asking for the bond angle of the $\rm C=C=O$ bond in $\rm H_2C=C=O$ . The VSEPR (valence shell electron pair repulsion) theory could help. Start by considering: how many electron domains are there on the carbon atom between these two bond?

Note that "electron domains" refer to covalent bonds and lone pairs collectively.

Each nonbonding pair (lone pair) of valence electrons counts as one electron domain.
Each covalent bond (single bond, double bond, or triple bond) counts as exactly one electron domain.

For example, in $\rm H_2C=C=O$ , the carbon atom at the center of that $\rm C=C=O$ bond has two electron domains:

This carbon atom has two double bonds: one $\rm C=C$ bond and one $\rm C=O$ bond. Even though these are both double bonds, in VSEPR theory, each of them count only as one electron domain.
Keep in mind that there are only four valence electrons in each carbon atom. It can be shown that all four valence electrons of this carbon atom are involved in bonding (two in each of the two double bonds.) Hence, there would be no nonbonding pair around this atom.

In VSEPR theory, electron domains around an atom repel each other. As a result, they would spread out (in three dimensions) as far away from each other as possible. When there are only two electron domains around an atom, the two electron domains would form a straight line- with one domain on each side of the central atom. (To visualize, consider the three atoms in this $\rm C=C=O$ bond as three spheres on a stick. The central $\rm C$ atom would be between the other $\rm C$ atom and the $\rm O$ atom.)

This linear geometry corresponds to a bond angle of $180^\circ$ .

3 0

3 years ago

Write the noble gas configuration for carbon

Ivanshal [37]

Answer:

[He] 2s2 2p2

Explanation:

6 0

4 years ago

After 15 minutes, 30 g of a sample of polonium-218 remain unchanged. If the original sample had a mass of 960 g, what is the hal

Keith_Richards [23]

Q1. The answer is 3 minutes.

Let's first calculate the remaining amount in percent:
If 960g is 100 percent (starting amount), 30 g is how many percents:
960 g : 100% = 30 g : x
x = 30 g * 100% / 960 g = 3.125% = 0.03125

Now, using the formula to calculate the number of half-lives:
(1/2)ⁿ = x,
where
x is the remaining amount: x = 0.03125
n is the number of half-lives
1/2 stands for half-life.

(1/2)ⁿ = 0.03125
⇒ n*log(1/2) = log(0.03125)
n = log(0.03125)/log(1/2) = log(0.03125)/log(0.5) = -1.505/-0.301 ≈ 5

The number of half-lives is 5.
Now, the number of half-lives (n) is a quotient of total time elapsed (T) and length of half-life (L):
n = T/L
We know:
n = 5
T = 15 min
L = ?

Thus
L = T/n
L = 15 min/5 = 3 minutes

Q2. Filtration should be chosen.
Filtration and distillation are used to separate mixtures - filtration for the separation of heterogeneous mixtures and distillation for the separation of homogeneous mixtures. Imagine that your pile of soil is a mixture. It consists of different components (the leaves, acorns, twigs), so it is the heterogeneous mixture. Therefore, to separate the leaves, acorns, twigs you will use filtration.

5 0

4 years ago

What is the difference between solvation and dissociation

mel-nik [20]

Answer: The key difference between dissociation and solvation is that dissociation is the breakdown of a substance into the atoms or ions from which the substance is made of whereas solvation is the dissolution of a substance in a solvent due to the attraction forces between the solvent molecules.

3 0

3 years ago