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sveta [45]
3 years ago
5

Within each group of four atoms or ions presented below. select the species that are isoelectronic with each other:Note: you are

not comparing the two columns, only selecting which within each column are isoelectronic with each other-not across the two columns.
Zn Ge²⁺
Fe³⁺ Cl⁻
Mn²⁺ B⁻
Ar C
Chemistry
1 answer:
Basile [38]3 years ago
5 0

Answer:

Fe³⁺ and Mn⁺²

B⁻ and C

Explanation:

Isoelectronic are species that have the same number of electrons. A neutral species has the same number of protons and electrons, and the number of protons in the atomic number (Z) found in the periodic table.

A cation (positive ions) lost the numbe of electrons indicate in its charge, and an anion (negative ions) gain the number of electrons indicate in its charge. So, let's identify the number of electrons (e-) in each one the atoms:

First column:

Zn: Z = 30, e- = 30

Fe³⁺: Z = 26, e- = 26 - 3 = 23

Mn⁺²: Z = 25, e- = 25 - 2 = 23

Ar: Z = 18

Isoelectronic: Fe³⁺ and Mn⁺²

Second column:

Ge⁺²: Z = 32, e- = 32-2 = 30

Cl⁻: Z = 17, e- = 17 + 1 = 18

B⁻: Z = 5, e- = 5 + 1 = 6

C: Z = 6, e- = 6

Isoelectronic B⁻ and C

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Since K is big, more product is expected.  This is because of mathematic principles.  A large numerator with a small denominator will produce a large number.

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For the solution resulting from dissolved 0.32 g of naphthalene (C10H8) in 25 g of benzene (C6H6) at temperature of 26.1°C, calc
grandymaker [24]

Answer:

See explanation

Explanation:

Number of moles of naphthalene = 0.32g/128.1705 g/mol = 0.0025 moles

Molality = number of moles/ mass of Solvent in kilograms

Molality = 0.0025/0.025 Kg

Morality = 0.1 m

But

∆T= K × i × m

Where ∆T = boiling point elevation

i= number of particles (this is equal to 1 because naphthalene is molecular and not ionic)

m= molality of naphthalene = 0.1 m

K= boiling point elevation constant = 5.12 °C/m

∆T= 5.12 °C/m ×0.1 = 0.512°C

For freezing point depression

∆T= K× i × m

Where ∆T= freezing point depression

i= number of particles (this is equal to 1 because naphthalene is molecular and not ionic)

m= molality of naphthalene = 0.1 m

K= freezing point depression constant = 2.67 °C/m

∆T= 2.67 °C/m ×0.1 = 0.267°C

From Raoult's law;

∆P = XBPA°

Where;

∆P = vapour pressure lowering

XB = mole fraction of solute

PA° = vapour pressure of pure solvent

Number of moles of solvent = mass/molar mass = 25g/ 78 g/mol= 0.3205 moles

Total number of moles = number of moles of solute + number of moles of solvent = 0.0025 moles + 0.3205 moles = 0.323 moles

Mole fraction of solute = 0.0025 moles/0.323 moles = 0.0077

Vapour pressure of benzene = 100 torr

Therefore;

∆P = 0.0077 × 100torr = 0.77 torr

Hence;

∆P = 0.77 torr

5 0
3 years ago
Given K = 3.61 at 45°C for the reaction A(g) + B(g) equilibrium reaction arrow C(g) and K = 7.19 at 45°C for the reaction 2 A(g)
Firlakuza [10]

Answer:

K = 0.55

Kp = 0.55

mol fraction B = 0.27

Explanation:

We need to calculate the equilibrium constant for the reaction:

C(g) + D(g) ⇄ 2B(g)              K₁= ?                       (1)

and we are given the following equilibria with their respective Ks

A(g) + B(g) ⇄ C(g)                 K₂= 3.61                 (2)

2 A(g) + D(g)  ⇄ C(g)             K₃= 7.19                 (3)

all at 45 ºC.

What we need to do to solve this question is to manipulate equations (2) and (3)  algebraically  to get our desired equilibrium (1).

We are allowed to reverse  reactions, in that case we take the reciprocal of K as our new K' ; we can also  add two equilibria together, and the new equilibrium constant will be the product of their respective Ks .

Finally if we multiply by a number then we raise the old constant to that factor to get the new equilibrium constant.

With all this  in mind, lets try to solve our question.

Notice A is not in our goal equilibrium (3)  and we want D as a reactant . That  suggests we should reverse the first equilibria and multiply it by two since we have 2 moles of B  as product in our  equilibrium (1) . Finally we would add (2) and (3) to get  (1) which is our final  goal.

2C(g)             ⇄  2A(g) + 2B(g)  K₂´= ( 1/ 3.61 )²  

                                   ₊

2 A(g) + D(g)  ⇄     C(g)               K₃ = 7.19  

<u>                                                                                    </u>

C(g) + D(g)     ⇄    2B(g)       K₁ = ( 1/ 3.61 )²   x  7.19

                                             K₁ = 0.55

Kp is the same as K = 0.55 since the equilibrium constant expression only involves  gases.

To compute the last part lets setup the following mnemonic  ICE table to determine the quantities at equilibrium:

pressure (atm)        C             D           B

initial                     1.64          1.64         0

change                    -x             -x        +2x

equilibrium          1.64-x         1.64-       2x

Thus since

Kp =0.55 = pB²/ (pC x pD) = (2x)²/ (1.64 -x)²  where p= partial pressure

Taking square root to both sides of the equation we have

√0.55 = 2x/(1.64 - x)

solving for x  we obtain a value of 0.44 atm.

Thus at equilibrium we have:

(1.64 - 0.44) atm = 1.20 atm = pC = p D

2(0.44) = 0.88 = pB

mole fraction of B = partial pressure of B divided into the total gas pressure:

X(B) = 0.88 / ( 1.20 + 1.20 + 0.88 ) = 0.27

8 0
3 years ago
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