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4 years ago

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For the following reaction, the reactants are favored at equilibrium. Classify each of the reactants and products based on their

strength as Bronsted-Lowry acids or bases. CH3COO- + H(CH3)3N+CH3COOH + (CH3)3N

1 answer:

Luden [163]4 years ago

8 0

Answer:

See explanation below

Explanation:

First, let's write again the reaction:

CH₃COO⁻ + H(CH₃)₃N⁺ <-----------> CH₃COOH + (CH₃)₃N

Now that the reaction is here, let's remember the basis of the bronsted - lowry theory:

An acid (HA) is a substance that can lose a proton (Hydrogen atom) to form a conjugate base. A base is a substance that accepts the proton (Hydrogen) and form a conjugate acid.

According to this definition, let's see the reaction again.

In the reactants, we see the CH3COO and the H(CH3)N. and the products are CH3COOH and (CH3)3N. The difference? well, we can see that the CH3COO now has a Hydrogen atom, this means that the CH3COO accepted the Hydrogen; this hydrogen was provided by the H(CH3)3N.

Therefore, the acid in this reaction is the H(CH₃)₃N⁺ and the conjugate base will be the (CH₃)₃N

The base in this reaction is the CH₃COO⁻ while the conjugate acid will be the CH₃COOH

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ammonium nitrate, which is prepared from nitric acid, is used as a nitrogen fertilizer. determine the percent composition of amm

Alina [70]

Answer:

Ammonium nitrate, (NH4NO3), a salt of ammonia and nitric acid, used widely in fertilizers and explosives. The commercial grade contains about 33.5 percent nitrogen, all of which is in forms utilizable by plants; it is the most common nitrogenous component of artificial fertilizers.

5 0

3 years ago

How much heat will be released when 0.750 moles of carbon monoxide (CO) react in an excess of oxygen gas? 2CO(g) + O2(g) → 2CO2(

lara31 [8.8K]

Answer:

A.566 kJ

B.424 kJ

C.212 kJ

D.755 kJ

Answer: C

Explanation:

3 0

4 years ago

Read 2 more answers

When 6.0 mol Al react with 13 mol HCl, what is the limiting reactant, and how many moles of H2 can be formed?

Ad libitum [116K]

You're looking for the number of moles of H2, and you have 6.0 mol Al and 13 mol HCL.

For the first part, you have to make your way from 6.0 mol of Al to mol of H2, right? For that to happen, you need to make a conversion factor that will cancel the mol Al, in such case use the 2 moles of Al from your equation to cancel them out. At the top of the equation, you can use the number of moles of H2 from the equation and find the moles that will be produced for the H2.

6.0mol Al x 3 mol H2/2 mol Al = 9 mol H2

For the second part, you have to make the same procedure, make a conversion factor that will cancel the mol of HCL and for that you need to use the 6 mol HCL from your equation, and at the numerator you can put the 3 mol of H2 from the equation so that you can find the number of moles of H2 that will be produced.

13 mol HCL x 3 mol H2/6 mol HCL = 6.5 mol H2

As it can be seen, HCL produces the less amount of H2 moles. Therefore, the reaction CANNOT produce more than 6.5 mol H2, in that case 6.5 mol will be the maximum number of moles that will be produced at the end because HCL does not have enough to produce more than 6.5 mol.

In that case HCL is the limiting reactant because it limits that will be produced, and so the answer is B!

6 0

3 years ago

A 3.00g of a certain Compound X, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of

andrew-mc [135]

Answer: The molecular formula for the given organic compound X is $C_6H_{8}O_7$

Explanation:

We are given:

Mass of $CO_2=4.13g$

Mass of $H_2O=1.13g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 4.13 g of carbon dioxide, = $\frac{12}{44}\times 4.13=1.13g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 1.13 g of water, $\frac{2}{18}\times 1.13=0.125g$ of hydrogen will be contained.

Mass of oxygen in the compound = (3.00) - (1.13+ 0.125) = 1.75 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.13g}{12g/mole}=0.094moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.125g}{1g/mole}=0.125moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{1.75g}{16g/mole}=0.109moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles

For Carbon = $\frac{0.094}{0.094}=1$

For Hydrogen = $\frac{0.125}{0.094}=1.33$

For Oxygen = $\frac{0.109}{0.094}=1.16$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 1: 1.33: 1.16

Converting them into whole number ratios by multiplying by 6:

The ratio of C : H : O = 6: 8: 7

Hence, the empirical formula for the given compound is $C_6H_8O_7$

Empirical mass = $6\times 12+8\times 1+7\times 16=192g$

The equation used to calculate the valency is :

$n=\frac{\text{molecular mass}}{\text{empirical mass}}$

Putting values in above equation, we get:

$n=\frac{192g/mol}{192g/mol}=1$

Multiplying this valency by the subscript of every element of empirical formula, we get:

$C_6H_8O_7\times 1=C_6H_{8}O_7$

Thus molecular formula for the given organic compound X is $C_6H_{8}O_7$

7 0

3 years ago

If 44.7 g of KCI (MM = 74.55 g/mol) are added to a 500.0 mL volumetric flask, and water is added to fill the flask, what is the

navik [9.2K]

1.199 M is the concentration of KCI in the resulting solution.

<h3 /><h3>What are moles?</h3>

A mole is defined as 6.02214076 × $10^{23}$ of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

No.of moles of KCI

$Moles = \frac{mass}{molar \;mass}$

$Moles = \frac{44.7 g}{74.5513 g/mol}$

= 0.599 moles

Vol.of the solution,V= 500 ml

= 0.5 liter

Molarity

$Molality = \frac{Moles \;solute}{Volume \;of \;solution \;in \;litre}$

$Molality = \frac{0.599 moles}{0.5 liter}$

= 1.199 M

Hence, 1.199 M is the concentration of KCI in the resulting solution.

Learn more about moles here:

brainly.com/question/8455949

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6 0

2 years ago