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3 years ago

Why is nitrobenzoic acid more acidic than ethanoic acid

Chemistry

1 answer:

IgorC [24]3 years ago

4 0

Explanation:

acidic strength increases if there is less electron density on a particular compound. and, as we know, greater the electron density on the compound,lesser acidic it will be.

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A 445N student eats 50g of Pringles. They feel guilty and decide to climb up some stairs to make up for it. How many flights of

krek1111 [17]

Answer:

808

Step-by-step explanation:

One serving of Original Pringles = 28 g = 150 Cal

Pringles ⟶ Cal 50 g × (150 Cal/28 g) = 268 Cal

Cal ⟶ J 268 Cal × (4184 J/1 Cal) = 1.121 × 10⁶ J

J ⟶ N·m 1.121 × 10⁶ J × (1 N·m/1 J) = 1.121 × 10⁶ N·m

N·m ⟶ m 1.126 × 10⁶ N·m/455N = 2460 m

m ⟶ km 2460 m × (1 km/1000 m) = 2.46 km

km ⟶ mi 2.46 km × (0.6214 mi/1 km) = 1.53 mi

mi ⟶ ft 1.53 mi ×(5280 ft/1 mi) = 8080 ft

ft ⟶ flights 8080 ft × (1 flight/10 ft) = 808 flights

The student would have to climb 808 flights of stairs to burn off the energy from 50 g of Pringles [that's equivalent to climbing the stairs in the Salesforce Tower (1070 ft) in San Francisco about 7½ times].

6 0

4 years ago

What is the ph at the equivalence point if it requires 32.5 mL of 0.15 M KOH to completely titrate 20 mL of a solution of propan

marusya05 [52]

Answer:

pH = 8.92

Explanation:

To solve this question we must know that the reaction of KOH with HC3H5O2 is:

KOH + HC3H5O2 → H2O + KC3H5O2

At equivalence point, all propanoic acid reacts to produce KC3H5O2.

This KC3H5O2 = C3H5O2⁻ is in equilibrium in water as follows:

C3H5O2⁻(aq) + H₂O(l) → OH⁻(aq) + HC3H5O2(aq)

<em>Where Kb = Kw / ka = 1x10⁻¹⁴/ 1.34x10⁻⁵ = 7.46x10⁻¹⁰</em>

is defined as:

Kb = 7.46x10⁻¹⁰ = [OH⁻] [HC3H5O2] / [C3H5O2⁻]

As both [OH⁻] [HC3H5O2] ions comes from the same equilibrium,

[OH⁻] = [HC3H5O2] = X

[C3H5O2⁻] is:

<em>Moles KOH = Moles </em>C3H5O2⁻:

0.0325L * (0.15mol / L) = 0.004875 moles

In 32.5 + 20mL = 52.5mL = 0.0525L:

0.004875 moles / 0.0525L = 0.09286M.

Replacing:

7.46x10⁻¹⁰ = [X] [X] / [0.09286M]

6.927x10⁻¹¹ = X²

X = 8.323x10⁻⁶M = [OH-]

As pOH = -log [OH-]

pOH = 5.08

pH = 14 -pOH

8 0

3 years ago

Consider the following reaction: NH4I(s) NH3(g) + HI(g) If a flask maintained at 674 K contains 0.138 moles of NH4I(s) in equili

xz_007 [3.2K]

Answer:

4.08 × 10⁻³

Explanation:

Step 1: Write the balanced reaction at equilibrium

NH₄I(s) ⇄ NH₃(g) + HI(g)

Step 2: Calculate the equilibrium constant

The equilibrium constant (K) is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. Only gases and aqueous species are included.

$K = [NH_3] \times [HI] = 4.34 \times 10^{-2} \times 9.39 \times 10^{-2} = 4.08 \times 10^{-3}$

4 0

4 years ago

Suppose the decomposition of dinitrogen monoxide proceeds by the following mechanism:

photoshop1234 [79]

Answer:

Part A: 2N₂O(g) ⇄ 2N₂(g) + O₂(g)

Part B: -r = K*[N₂O]²

Part C: K= k1*k2

Explanation:

Part A

To do the balance chemical question for the overall chemical reaction, we must sum the reaction of the steps, eliminating the intermediaries, which are the compounds that have the same amount both at reactants and products (bolded).

N₂O(g) ⇄ N₂(g) + O(g)

N₂O(g) + O(g) ⇄ N₂(g) + O₂(g)

---------------------------------------------

2N₂O(g) + O(g) ⇄ 2N₂(g) + O(g) + O₂(g)

2N₂O(g) ⇄ 2N₂(g) + O₂(g)

Part B

The velocity of the reaction (r) can be calculated based on the reactants or based on the products. Let's do it based on the disappearing of the reactant. Because it is disappearing, the variation at its concentration must be negative, so the rate will be negative.

Let's suppose its an elementary reaction, so, the concentration of the reactant must be elevated by its coefficient. And let's call the overall rate constant as K:

-r = K*[N₂O]²

Part C

Because the steps were summed, and the reactions were not multiplied by a constant or inverted, the constant K is just the multiplication of the constants of the steps:

K= k1*k2

8 0

4 years ago

The half-life of a first-order reaction is 13 min. If the initial concentration of reactant is 0.13 M, it takes ________ min for

Zigmanuir [339]

Answer:

Therefore it takes 8.0 mins for it to decrease to 0.085 M

Explanation:

First order reaction: The rate of reaction is proportional to the concentration of reactant of power one is called first order reaction.

A→ product

Let the concentration of A = [A]

$\textrm{rate of reaction}=-\frac{d[A]}{dt} =k[A]$