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The decomposition of HBr(g) into elemental species is found to have a rate constant of 4.2 ×10−3atm s−1. If 2.00 atm of HBr are

Dennis_Churaev [7]

Answer:

7,94 minutes

Explanation:

If the descomposition of HBr(gr) into elemental species have a rate constant, then this reaction belongs to a zero-order reaction kinetics, where the reaction rate does not depend on the concentration of the reactants.

For the zero-order reactions, concentration-time equation can be written as follows:

[A] = - Kt + [Ao]

where:

[A]: concentration of the reactant A at the t time,
[A]o: initial concentration of the reactant A,
K: rate constant,
t: elapsed time of the reaction

To solve the problem, we just replace our data in the concentration-time equation, and we clear the value of t.

Data:

K = 4.2 ×10−3atm/s,

[A]o=[HBr]o= 2 atm,

[A]=[HBr]=0 atm (all HBr(g) is gone)

We clear the incognita :

[A] = - Kt + [Ao]............. Kt = [Ao] - [A]

t = ([Ao] - [A])/K

We replace the numerical values:

t = (2 atm - 0 atm)/4.2 ×10−3atm/s = 476,19 s = 7,94 minutes

So, we need 7,94 minutes to achieve complete conversion into elements ([HBr]=0).

6 0

3 years ago

If 10.62 mL of a standard 0.3330 M KOH solution reacts with 98.20 mL of CH3COOH solution, what is the molarity of the acid solut

Nikolay [14]

Answer:

0.036 M of $CH_{3} COOH$

Explanation:

It is an example of acid-base neutralization reaction.

KOH + $CH_{3} COOH$ ----> $CH_{3} COO^{-} K^{+}$ + $H_{2}O$

Base Acid Salt

When two component react then the number of moles of both the component should be same, therefore the number of moles and acids and bases should be the same in the following .

Molarity= $\frac{\textrm{No. of Moles}}{\textrm{Volume of the Particular Solution}}$

No.of moles= Molarity × Volume of the Particular Solution

Therefore,

$M_{1}V_{1} =M_{2}V_{2}$ ------------------------------(1)

where

$M_{1}$ = Molarity of Acid

$V_{1}$ = Volume of Acid

$M_{2}$ = Molarity of Base

$V_{2}$ = Volume of Base

$M_{1}$ =0.3330 M

$V_{1}$ =10.62 mL

$V_{2}$ =98.2 mL

$M_{2}$ =??(in M)

Plugging in Equation 1,

0.3330 × 10.62 = $M_{2}$ × 98.2

$M_{2}$ = $\frac{0.3330*10.62}{98.2}$

$M_{2}$ =0.036 M

3 0

3 years ago

For alkyl halides used in SN1 and SN2 mechanisms, rank the leaving groups in order of reaction rate. You are currently in a rank

Alex777 [14]

Answer:

Iodide> Bromide > chloride > flouride

Explanation:

During a nucleophilic substitution reaction, a nucleophilie replaces another in a molecule.

This process may occur via an ionic mechanism (SN1) or via a concerted mechanism (SN2).

In either case, the ease of departure of the leaving group is determined by the nature of the C-X bond. The stronger the C-X bond, the worse the leaving group will be in nucleophilic substitution. The order of strength of C-X bond is F>Cl>Br>I.

Hence, iodine displays the weakest C-X bond strength and it is thus, a very good leaving group in nucleophillic substitution while fluorine displays a very high C-X bond strength hence it is a bad leaving group in nucleophilic substitution.

Therefore, the ease of the use of halide ions as leaving groups follows the trend; Iodide> Bromide > chloride > flouride

4 0

3 years ago

What is the [OH-] if the [H3O+] is 1 x 10 -6?

Mariulka [41]

The answer is [OH⁻] = 1 x 10⁻⁸.

To find OH⁻, divide the ionic product of water by [H₃O⁺] as :

OH⁻ + H₃O⁺ = H₂O

[OH⁻] = 1 x 10⁻¹⁴ / 1 x 10⁻⁶
[OH⁻] = 1 x 10⁻⁸

5 0

2 years ago

The number of B2H6 molecules

zlopas [31]

Answer:

6.022 × 10²³ molecules of B₂H₆

Explanation:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

18 g of water = 1 mole = 6.022 × 10²³ molecules of water

1.008 g of hydrogen = 1 mole = 6.022 × 10²³ atoms of hydrogen

B₂H₆ molecules:

21.63 g = one mole of B₂H₆= 6.022 × 10²³ molecules of B₂H₆

8 0

3 years ago