Which of the following is the greatest speed: 20 m/s, 200 cm/s, or 0.2 km/ s

Methane, CH4, reacts with I2 according to the reaction CH4(g)+I2(g)⇌CH3I(g)+HI(g)

gtnhenbr [62]

Answer:

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

Explanation:

Kp is the equilibrium constant for the partial pressure of the gases in the reaction, and it is calculated for a general equation:

aA(g) + bB(g) ⇄ cC(g) + dD(g)

$Kp = \frac{(pC)^cx(pD)^d}{(pA)^ax(pB)^b}$ , where p is the partial pressure in the equilibrium. By the reaction given:

CH₄(g) + I₂(g) ⇄ CH₃I(g) + HI(g)

105.1 torr 7.96 torr 0 0 initial partial pressure

-x -x +x +x react

105.1-x 7.96-x x x equilibrium

Then:

$Kp = \frac{pCH3IxpHI}{pCH4xpI2} = \frac{x^2}{(105.1-x)(7.96-x)}$

$2.26x10^{-4} = \frac{x^2}{836.596 - 113.06x -x^2}$

x² = 0.1891 - 0.0255x -2.26x10⁻⁴x²

0.9997x² + 0.0255x - 0.1891 = 0

Using Bhaskara's rule:

Δ = (0.0255)² - 4x(0.9997)x(-0.1891)

Δ = 0.7568

$x = \frac{-b+/-\sqrt{0.7568} }{2a} = \frac{-0.0255 +/-0.8699}{1.9994}$

Using only the positive term, x = 0.42 torr.

So,

pCH₄ = 105.1 - 0.42 = 104.68 torr

pI₂ = 7.96 -0.42 = 7.54 torr

pCH₃I = 0.42 torr

pHI = 0.42 torr

8 0

2 years ago

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Place each charge form of alanine under the pH condition where it would be the predominant form. The pKa values for the carboxyl

damaskus [11]

Answer:

(A) pH < 1 the predominant form is the cation: H3C-C(H)(NH3+)-COOH

(B) pH = pl the predominant form is the zwitterion H3C-C(H)(NH3+)-COO-

(C) pH > 11 the predominant form is the anion: H3C-C(H)(NH2)-COO-

(D) Does not occurs in any significant pH: H3C-C(H)(NH2)-COOH

Explanation:

Amino acids are bifunctional because they have an amine group and a carboxyl group. The amine group is a weak base and the carboxyl group is a weak acid, but the pKa of both groups will depend on the whole structure of the amino acid. Also, every amino acid has an isoelectric point (pI), which means the pH were the predominant form of the amino acid is the zwitterion. The structure of the alanine (CH3CH2NH2COOH) shows it has the carboxyl group at C1 with a pKa1 of 2.3 and the amino group at C2 whit the pKa2 of 9.7. The isoelectric poin (pI) of Alanine is 6. Consequently, the protonation of the molecule will depend on the pH of the solution. There are three possibilities:

1) If the pH is under the pKa of the carboxyl group (2.3) the predominant form will be with the amino group protonated, forming a cation (CH3CH(NH3+)COOH).

2) If the pH is between pKa1 (2.3) and pKa2 (9.7) the predominant form will be the zwitterion (CH3CH(NH3+)(COO-)).

3) If the pH is upper the pKa2 of the amino group (9.7) the predominant form will be with the carboxyl group deprotonated, forming an anion (CH3CHNH2(COO-)).

8 0

2 years ago

Compute the end-to-end separation (direct distance between molecule ends), in Angstroms, of a coiled PVC molecule with a number-

MArishka [77]

Answer:

5156 Â

Explanation:

Chains of monomers that are being linked together constitute what is called polymers. in this question; we are to compute the end to end separation (direct distance between molecule ends), in Angstroms, of a coiled PVC molecule.

Using the following approach:

The first step is to determine the repeating units in polymer

The number of repeating unit = average molecular weight/ atomic weight of PVC

Given that:

average molecular weight = 256,131 g/mol

atomic weight of PVC = 62.498 g

Then;

The number of repeating unit = 256,131 g/mol / 62.498 g = 4098.227143 mol

Now the distance is calculated by using the formula:

d = (C-C) × sin 109.5/2

C-C bond length = 1.54 angstroms

tetrahedral bond angle = 109.5°

Then d = (1.54) × sin 109.5/2

d = 1.258

Thus ; the end to end separation is :

4098.227143 × 1.258 = 5155.57 Â

The answer is 5156 Â (since no decimal should be included )

6 0

2 years ago

A student is filtering a mixture of sand and salt water into a beaker. What will be found in the beaker after the filtration is

Sergio [31]

The student would find the water and sand, because salt dissolves in water unless it was ocean water or sea water

3 0

2 years ago

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Aluminum–lithium (Al–Li) alloys have been developed by the aircraft industry to reduce the weight and improve the performance of

gtnhenbr [62]

Answer:

The concentration of Li (in wt%) is 3,47g/mol

Explanation:

To obtain the 2,42g/cm³ of density:

2,42g/cm³ = 2,71g/cm³X + 0,534g/cm³Y (1)

Where X is molar fraction of Al and Y is molar fraction of Li.

X + Y = 1 (2)

Replacing (2) in (1):

Y = 0,13

Thus, X = 0,87

The weight of Al and Li is:

0,87*26,98g/mol = 23,4726 g of aluminium

0,13*6,941g/mol = 0,84383 g of lithium

The concentration of Li (in wt%) is:

0,84383g/(0,84383g+23,4726g) ×100= 3,47%

6 0

1 year ago