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2 years ago

Which side of this proton transfer reaction is favored at equilibrium?

Chemistry

1 answer:

Maru [420]2 years ago

3 0

The left side proton transfer will be favored and the reactants will convert into products , Option C is the right answer.

<h3>What is Proton Transfer ?</h3>

A step in a process in which the proton is removed from one species and accepted by another species is called Proton Transfer.

It generally takes place between an acid and a base.

At equilibrium the reactants are mostly converted into products , At equilibrium the equilibrium lie towards the right side .

Therefore left side proton transfer will be favored and the reactants will convert into products.

To know more about Proton Transfer

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Help please!

umka2103 [35]

Answer: In factories they burn coal under a giant container(a really big gumbo pot pretty much). when the fire is hot enough it starts to boil. When it boils it creates steam and the steam blows on a fan that makes electricity.

4 0

3 years ago

A 14.3-cm3 sample of tin has a mass of 0.104 kg.

VladimirAG [237]

This is a true statement if it is density you are looking for... Density problem.....

Density is the ratio of the mass of an object to its volume.
D = m / V
D = 104g / 14.3 cm³ = 7.27 g/cm³ .............. to three significant digits

The conventions for the units of density is that grams per cubic centimeter (g/cm³) are usually used for solids, but will work for anything. Grams per milliliter (g/mL) are usually used for liquids and grams per liter (g/L) are for gases. Therefore, by convention, the units for tin (a solid) should be in grams per cubic centimeter.

Since 1 mL is equivalent to 1 cm³, then the density could be expressed as 7.27 g/mL.

The accepted value for the density of tin is 7.31 g/cm³

7 0

3 years ago

Glucose, C6H12O6,C6H12O6, is used as an energy source by the human body. The overall reaction in the body is described by the eq

Ilia_Sergeevich [38]

Answer:

Mass of oxygen = 61.824 g

Mass of carbon dioxide = 85.01 g

Explanation:

Given data:

Mass of glucose = 58 g

Mass of carbon dioxide = ?

Mass of oxygen = ?

Solution:

First of all we will write the balanced chemical equation,

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

Moles of glucose:

Number of moles = mass / molar mass

Number of moles = 58 g/180 g/mol

Number of moles = 0.322 mol

Now we compare the moles of oxygen with glucose from balance chemical equation.

C₆H₁₂O₆ : O₂

1 : 6

0.322 : 0.322×6 = 1.932 mol

Mass of oxygen:

Mass of oxygen = number of moles × molar mass

Mass of oxygen = 1.932 mol × 32 g/mol

Mass of oxygen = 61.824 g

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

C₆H₁₂O₆ : CO₂

1 : 6

0.322 : 0.322×6 = 1.932 mol

O₂ : CO₂

6 : 6

1.932 : 1.932

Mass of carbon dioxide;

mass of carbon dioxide = number of moles × molar mass

mass of carbon dioxide = 1.932 mol × 44 g/mol

mass of carbon dioxide = 85.01 g

7 0

3 years ago

Manganese(iv) oxide reacts with aluminum to form elemental manganese and aluminum oxide: 3mno2+4al→3mn+2al2o3part awhat mass of

Oxana [17]

<span>12.4 g First, calculate the molar masses by looking up the atomic weights of all involved elements. Atomic weight manganese = 54.938044 Atomic weight oxygen = 15.999 Atomic weight aluminium = 26.981539 Molar mass MnO2 = 54.938044 + 2 * 15.999 = 86.936044 g/mol Now determine the number of moles of MnO2 we have 30.0 g / 86.936044 g/mol = 0.345081265 mol Looking at the balanced equation 3MnO2+4Alâ†’3Mn+2Al2O3 it's obvious that for every 3 moles of MnO2, it takes 4 moles of Al. So 0.345081265 mol / 3 * 4 = 0.460108353 mol So we need 0.460108353 moles of Al to perform the reaction. Now multiply by the atomic weight of aluminum. 0.460108353 mol * 26.981539 g/mol = 12.41443146 g Finally, round to 3 significant figures, giving 12.4 g</span>

7 0

3 years ago

The reaction is proceeding at a rate of 0.0080 Ms-1 in 50.0 mL of solution in a system with unknown concentrations of A and B. W

Leokris [45]

Answer:

0.0010 mol·L⁻¹s⁻¹

Explanation:

Assume the rate law is

rate = k[A][B]²

If you are comparing two rates,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \dfrac{k_{2}\text{[A]}_2[\text{B]}_{2}^{2}}{k_{1}\text{[A]}_1[\text{B]}_{1}^{2}}= \left (\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}}\right ) \left (\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}\right )^{2}$

You are cutting each concentration in half, so

$\dfrac{\text{[A]}_{2}}{\text{[A]}_{1}} = \dfrac{1}{2}\text{ and }\dfrac{\text{[B]}_{2}}{\text{[B]}_{1}}= \dfrac{1}{2}$

Then,

$\dfrac{\text{rate}_{2}}{\text{rate}_{1}} = \left (\dfrac{1}{2}\right ) \left (\dfrac{1}{2}\right )^{2} = \dfrac{1}{2}\times\dfrac{1}{4} = \dfrac{1}{8}\\\\\text{rate}_{2} = \dfrac{1}{8}\times \text{rate}_{1}= \dfrac{1}{8}\times \text{0.0080 mol$\cdot$L$^{-1}$s$^{-1}$} = \textbf{0.0010 mol$\cdot$L$^{-1}$s$^{-1}$}\\\\\text{The new rate is $\large \boxed{\textbf{0.0010 mol$\cdot$L$^\mathbf{{-1}}$s$^{\mathbf{-1}}$}}$}$

8 0

2 years ago