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2 years ago

Identify the most and the least acidic compound in each of the following sets.

Chemistry

1 answer:

il63 [147K]2 years ago

3 0

Answer:

See explanation

Explanation:

Our answer options for this question are:

a. 2-chlorobutanoic acid:_______ 2-chlorobutanoic acid:_______ 3-chlorobutanoic acid:______.

b. 2,4-dinitrobenzoic acid:______ p-nitrobenzoic acid:______ p-bromobenzoic acid:_______.

c. p-cyanobenzoic acid:________ benzoic acid:_______ p-aminobenzoic acid:______

We have to check each set of molecules

a. 2-chlorobutanoic acid, 3-chlorobutanoic acid

In this case, the difference between these molecules is the position of "Cl". If the chlorine atom is closer to the acid group, we will have a higher inductive effect. So, the bond O-H would be weaker and we will have more acidity. So, the molecule with more acidity is 2-chlorobutanoic acid and the less acidic would be 3-chlorobutanoic acid.

b. 2,4-dinitrobenzoic acid, p-nitrobenzoic acid, p-bromobenzoic acid

In this case, we have several structural differences. In all the structure, we have deactivating groups ( $Br$ and $NO_2$ ). If we have a deactivating group the acidity will increase. In the case of "Br", we have a weak deactivating, so, this will be the less acidic one (p-bromobenzoic acid)

in 2,4-dinitrobenzoic acid we have two deactivating groups, therefore, this would be the most acid compound.

c. p-cyanobenzoic acid, benzoic acid, p-aminobenzoic acid

On these molecules, we have several structural differences. In p-cyanobenzoic acid we have a deactivating group, therefore in this molecule we will have more acidity. In the p-aminobenzoic acid, we have an activating group, so, this would be the less acidic compound.

See figure 1

I hope it helps!

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Calculate ΔG o for the following reaction at 25°C: 3Mg(s) + 2Al3+(aq) ⇌ 3Mg2+(aq) + 2Al(s) Enter your answer in scientific notat

larisa [96]

Answer:

-3.7771 × 10² kJ/mol

Explanation:

Let's consider the following equation.

3 Mg(s) + 2 Al³⁺(aq) ⇌ 3 Mg²⁺(aq) + 2 Al(s)

We can calculate the standard Gibbs free energy (ΔG°) using the following expression.

ΔG° = ∑np . ΔG°f(p) - ∑nr . ΔG°f(r)

where,

n: moles

ΔG°f(): standard Gibbs free energy of formation

p: products

r: reactants

ΔG° = 3 mol × ΔG°f(Mg²⁺(aq)) + 2 mol × ΔG°f(Al(s)) - 3 mol × ΔG°f(Mg(s)) - 2 mol × ΔG°f(Al³⁺(aq))

ΔG° = 3 mol × (-456.35 kJ/mol) + 2 mol × 0 kJ/mol - 3 mol × 0 kJ/mol - 2 mol × (-495.67 kJ/mol)

ΔG° = -377.71 kJ = -3.7771 × 10² kJ

This is the standard Gibbs free energy per mole of reaction.

5 0

2 years ago

What mass of AgBr can be produced starting with 34.3G of NaBr? (Stoichiometry method)

olga nikolaevna [1]

The equation is already balanced so the mole ratio is just 1/1 and from the question you could conclude that the equation is a single displacement

5 0

2 years ago

A gas occupies 22.4 l at stp and 14.5 l at 100c and 2.00 atm pressure. how many moles of gas did the system gain or lose?

azamat

At standard temperature and pressure 22.4 l of an ideal gas would contain 1 mole. in order to find the change in moles we must look at the ideal gas law PV=nRT where P=Pressure V=volume n=Moles R= Gas constant T= Temperature. To simplify this equation we will be using the gas constant at .08206 L-atm/mol-K. We must first convert 100c to k which is 373.15. Then we can plug the values into our equation which gives us (2atm)(14.5 l)=(n)(.08206 L-atm/mol-K)(373.15). After some basic algebra we get the moles to equal roughly .95 which is .05 moles less than our original system.

6 0

2 years ago

A poisoned pill contains 0.00048 moles of KCN. How many molecules are in this sample?

mario62 [17]

Answer:

$2.89 \times {10}^{20} \: \: molecules$

Explanation:

The number of molecules of KCN can be found by using the formula

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

N = 0.00048 × 6.02 × 10²³

We have the final answer as

$2.89 \times {10}^{20} \: \: \: molecules$

Hope this helps you

7 0

1 year ago

3. According to the label on a bottle of concentrated hydrochloric acid, the contents are 36.0% HCl by mass and have a density o

velikii [3]

Answer:

a) 11.64 M

b) 43 mL

c) 1.7 kg

Explanation:

a) Let's use a basis of the calculus of 1000 mL (1 L) of the concentrated solution. If the solution has 1.18 g/mL, it has:

1.18*1000 = 1180 g.

The mass of HCl will be then:

mHCl = 1180*0.36 = 424.8 g

The molar mass of HCl is 36.5 g/mol, so the number of moles is the mass divided by the molar mass:

nHCl = 424.8/36.5 = 11.64 mol

The molarity is the number of moles divided by the volume in L:

Molarity = 11.64 M

b) To prepare a solution by dilution of a concentrated one, we can use the equation:

C1V1 = C2V2

Where C is the concentration, V is the volume, 1 is the concentrated solution, and 2 the final solution. So:

11.64*V1 = 2.00*0.250

V1 = 0.0429 L ≅ 43 mL

c) The neutralization will happen by the equation:

HCl + NaHCO₃ → NaCl + CO₂ + H₂O

So, 1 mol of NaHCO₃ is needed to react with 1 mol of HCl. At 1.75 L, the number of moles of the acid is:

nHCl = 1.75*11.64 = 20.37 mol

The molar mass of NaHCO₃ is 84 g/mol so the mass needed is the molar mass multiplied by the number of moles:

m = 84*20.37 = 1,711.08 g

m = 1.7 kg

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2 years ago