Be sure to answer all parts. (a) How many atoms are directly bonded to the central atom in a trigonal planar molecule?

i. two
ii. three
iii. six
iv. eight

(b) How many atoms are directly bonded to the central atom in a trigonal bipyramidal molecule?

i. three
ii. four
iii. five
iv. six

(c) How many atoms are directly bonded to the central atom in an octahedral molecule?

i. three
ii. four
iii. six
iv. eight

Answer:

c. 6,3x10⁻¹¹M

Explanation:

The solubility of a buffer is defined as the concentration of the dissolved solid in a saturated solution. For the Cd(OH)₂, solubility is:

[Cd²⁺] = S

The dissolution of Cd(OH)₂ is:

Cd(OH)₂ ⇄ Cd²⁺ + 2OH⁻

And the ksp is defined as:

ksp = [Cd²⁺][OH⁻]²

As ksp = 2,5x10⁻¹⁴ and [OH⁻] at pH=12,30 = 10^-(14-12,30) = 0,01995M

2,5x10⁻¹⁴ = [Cd²⁺]×(0,01995M)²

[Cd²⁺] = 6,3x10⁻¹¹M

That means solubility is c. 6,3x10⁻¹¹M

I hope it helps!

Final answer:

The molar solubility of Cd(OH)2 when buffered at a pH of 12.30 can be calculated using the concept of hydrolysis. The correct answer is 6.3 x 10^(-11) M.

Explanation:

To calculate the molar solubility of Cd(OH)2 when buffered at a pH of 12.30, we need to use the concept of hydrolysis. Cd(OH)2 is a slightly soluble salt that undergoes hydrolysis in aqueous solution. At a high pH value, OH- ions react with water to form more OH- ions, shifting the equilibrium towards the hydrolysis reaction.

1. First, we write the balanced equation for the hydrolysis reaction: Cd(OH)2(s) ⇌ Cd2+(aq) + 2OH-(aq)
2. Since OH- is being produced, we can assume that the concentration of OH- is much greater than that of Cd2+. Therefore, we can ignore the concentration of Cd2+ when calculating the solubility product (Ksp).
3. Next, we use the equation for the hydrolysis reaction to write the expression for the solubility product constant (Ksp): Ksp = [Cd2+][OH-]^2
4. The concentration of OH- ions in a basic solution is related to the pH by the equation: pOH = 14 - pH
5. Using this equation, we can calculate the pOH of the buffered solution: pOH = 14 - 12.30 = 1.70
6. Then, we convert the pOH back to OH- concentration: [OH-] = 10^(-pOH) = 10^(-1.70)
7. Finally, we substitute the calculated [OH-] into the expression for Ksp to solve for the molar solubility of Cd(OH)2: [Cd(OH)2] = sqrt(Ksp / [OH-]^2)

After performing the calculations, the molar solubility of Cd(OH)2 when buffered at a pH of 12.30 is approximately 6.3 x 10^(-11) M. Therefore, the correct answer is option c. 6.3 x 10^(-11) M.

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The elements according to the decreasing atomic radius are arranged as-

K, Ca, Sc, Ga, Ge, As, Br, Kr

An atomic radius is half the distance between adjacent atoms of the same element in a molecule. It is a measure of the size of the element’s atoms, which is typically the mean distance from the nucleus centre to the boundary of its surrounding shells of the electrons.

An atom gets larger as the number of electronic shells increase; therefore the radius of atoms increases as you go down a certain group in the periodic table of elements. The atomic radius decreases on moving from left to right across a period.

Thus the elements according to the decreasing atomic radius are arranged as -

K, Ca, Sc, Ga, Ge, As, Br, Kr

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Answer:

Moles of hydrogen formed = 3.5 moles

Explanation:

Given that:-

Moles of aluminium= 4.0 mol

Moles of hydrogen bromide = 7.0 mol

According to the reaction:-

2 moles of aluminum react with 6 moles of hydrogen bromide

1 mole of aluminum react with 6/2 moles of hydrogen bromide

4 moles of aluminum react with (6/2)*4 moles of hydrogen bromide

Moles of hydrogen bromide = 12 moles

Available moles of hydrogen bromide = 7.0 moles

Limiting reagent is the one which is present in small amount. Thus, hydrogen bromide is limiting reagent. (7.0 < 12)

The formation of the product is governed by the limiting reagent. So,

6 moles of hydrogen bromide on reaction forms 3 moles of hydrogen

1 moles of hydrogen bromide on reaction forms 3/6 moles of hydrogen

7 moles of hydrogen bromide on reaction forms (3/6)*7 moles of hydrogen

Moles of hydrogen formed = 3.5 moles

Answer:

3.5 mol H2, HBr (limiting reactant)

Explanation:

4.0 mol Al × 3 mol H2/ 2 mol Al = 6.0 mol H2

7.0 mol HB ×3 mol H2/ 6mol HBr = 3.5 mol H2

Since 7.0mol of HBr will produce less H2 than 4.0mol of Al, HBr will be the limiting reactant, and the reaction will produce 3.5mol of H2.

Answer:

38.96383282 amu

Explanation:

39.0983 = (40.9618 0.067302) + ( ? (1-0.067302)

39.0983 = 2.756811064 + ( ? 0.932698)

subtract 2.756811064 from both sides

36.34148894 = ( ? 0.932698)

divide both sides by 0.932698

? = 38.96383282 amu

Answer:

38.96383282 amu

Explanation:

39.0983 = (40.9618  0.067302) + ( ?  (1-0.067302)

39.0983 = 2.756811064 + ( ?  0.932698)

subtract 2.756811064 from both sides

36.34148894 = ( ?  0.932698)

divide both sides by 0.932698

? = 38.96383282 amu

Answer:

ΔHrxn = 193107.69 J/mol

Explanation:

ΔHrxn = mcΔT

m = mass

c = heat capacity

ΔT = temperature variation

density = m/V

m = density x V

m = 1.00 g/mL x 400.0 mL

m = 400.0 g

ΔHrxn = mcΔT

ΔHrxn = 400 g x 4.184 J/g°C x 6.00 °C

ΔHrxn = 10041.6 J

CaO + 2HCl  → CaCl₂ + H₂O

CaO = 56.0774 g/mol

2.90 g CaO = 0.052 mol

400.0 mL of 1.500 mol/L HCl = 0.6 mol HCl

ΔHrxn = 10041.6 J is for 0.052 mol of CaO

ΔHrxn = 193107.69 J is for 1 mol of CaO

Answer:

it’s electron configuration is 1s^2 2s^2 2p^4. To determine valence electrons, add the outermost s and p orbitals. In an oxygen atom, 8 electrons are present. Electron present in the first shell (n=1) 2n^2=2 (1)^2=2 (1)=2.