According to the Brensted-Lowry model write the equation for provided acid showing

An acetylide anion is more stable than either a vinylic or alkyl anion due to Group of answer choices the greater stability of a

sashaice [31]

Answer:

the stabilization of the negative charge in orbitals with higher s character

Explanation:

Acetylide anion is a carbon anion compound or popularly called carbanion. Now Acetylide anion is sp hybridized. However acetylide anion tends to be more acidic as we move from sp³ to sp, hence acidicity increases, which makes sp to have the highest acidity and become the most stable.

So, we can conclude that the acetylide anion is more acidic due to the stabilization of the negative charge in orbitals with higher s character and as the s character increases, acidic nature of acetylide anion also increases.

5 0

2 years ago

Equal molar quantities of Ca2 and EDTA (H4Y) are added to make a 0.010 M solution of CaY2- at pH 10. The formation constant for

abruzzese [7]

Answer:

the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

Explanation:

Given the data in the question;

$Ca^{2+$ + $y^{4-$ ⇄ $CaY^{2-$

Formation constant Kf

Kf = $CaY^{2-$ / ( [ $Ca^{2+$ ][ $y^{4-$ ] ) = 5.0 × 10¹⁰

Now,

[ $y^{4-$ ] = $\alpha _4CH_4Y$ ; ∝₄ = 0.35

so the equilibrium is;

$Ca^{2+$ + $H_4Y$ ⇄ $CaY^{2-$ + 4H⁺

Given that; $CH_4Y$ = $Ca^{2+$ { 1 mol $Ca^{2+$ reacts with 1 mol $H_4Y$ }

so at equilibrium, $CH_4Y$ = $Ca^{2+$ = x

∴

$Ca^{2+$ + $y^{4-$ ⇄ $CaY^{2-$

x + x 0.010-x

since Kf is high, them x will be small so, 0.010-x is approximately 0.010

so;

Kf = $CaY^{2-$ / ( [ $Ca^{2+$ ][ $y^{4-$ ] ) = $CaY^{2-$ / ( [ $Ca^{2+$ ][ $\alpha _4CH_4Y$ ] ) = 5.0 × 10¹⁰

⇒ $CaY^{2-$ / ( [ $Ca^{2+$ ][ $\alpha _4CH_4Y$ ] ) = 5.0 × 10¹⁰

⇒ 0.010 / ( [x][ 0.35 × x] ) = 5.0 × 10¹⁰

⇒ 0.010 / 0.35x² = 5.0 × 10¹⁰

⇒ x² = 0.010 / ( 0.35 × 5.0 × 10¹⁰ )

⇒ x² = 0.010 / 1.75 × 10¹⁰

⇒ x² = 5.7142857 × 10⁻¹³

⇒ x = √(5.7142857 × 10⁻¹³)

⇒ x = 7.559 × 10⁻⁷

Therefore, the concentration of free Ca2⁺ in this solution is 7.559 × 10⁻⁷

8 0

2 years ago

2Al + 6HCl → 2AlCl3 + 3H2 If 85.0 grams of HCl react, how many moles of H2 are produced?

Murrr4er [49]

Answer:

1.17 mol

Explanation:

Step 1: Write the balanced equation

2 Al + 6 HCl → 2 AlCl₃ + 3 H₂

Step 2: Calculate the moles corresponding to 85.0 g of HCl

The molar mass of HCl is 36.46 g/mol.

85.0 g × 1 mol/36.46 g = 2.33 mol

Step 3: Calculate the number of moles of H₂ produced from 2.33 moles of HCl

The molar ratio of HCl to H₂ is 6:3.

2.33 mol HCl × 3 mol H₂/6 mol H₂ = 1.17 mol H₂

8 0

2 years ago

Complete the mechanism for the reaction of 2-butanol in sulfuric acid at 140 °C by adding any missing atoms, bonds, charges, non

Svetlanka [38]

Answer:

2-Butene

Explanation:

The first step is the ionization of the acid to produce the hydronium ion. Then the OH will attack this ion to produce a charged species that can be stabilized when H2O is produced.

Then an elimination takes place to produce the more substituted alkene 2-butene and the hydronium ion is gain produced.

4 0

3 years ago

if you started with 2.34 grams of methane and 8.32 grams of oxeygyn and the combustion reaction went to completion how many gram

Romashka-Z-Leto [24]

Answer:

Mass of CO₂ produced = 5.72 g

Explanation:

Given data:

Mass of methane = 2.34 g

Mass of oxygen = 8.32 g

Mass of CO₂ produced = ?

Solution:

Chemical equation:

CH₄ + 2O₂ → CO₂ + 2H₂O

Number of moles of methane:

Number of moles = mass/molar mass

Number of moles = 2.34 g/ 16 g/mol

Number of moles = 0.146 mol

Number of moles of oxygen:

Number of moles = mass/molar mass

Number of moles = 8.32 g/ 32 g/mol

Number of moles = 0.26 mol

Now we will compare the moles of carbon dioxide with oxygen and methane.

CH₄ : CO₂

1 : 1

0.146 : 0.146

O₂ : CO₂

2 : 1

0.26 : 1/2×0.26 = 0.13 mol

Less number of moles of CO₂ are produced by oxygen thus oxygen will react as limiting reactant.

Mass of CO₂:

Mass = number of moles × molar mass

Mass = 0.13 mol × 44 g/mol

Mass = 5.72 g

4 0

2 years ago