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zhannawk [14.2K]
4 years ago
13

How can you make a solution saturated

Chemistry
1 answer:
Vlada [557]4 years ago
3 0
You add solute until it can't dissolve anymore.
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Given the following reaction and data, A + B → Products
natima [27]

Answer:

a. Rate = k×[A]

b. k = 0.213s⁻¹

Explanation:

a. When you are studying the kinetics of a reaction such as:

A + B → Products.

General rate law must be like:

Rate = k×[A]ᵃ[B]ᵇ

You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.

If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1

Rate = k×[A]¹[B]ᵇ

In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]

Rate = k×[A][B]⁰

<h3>Rate = k×[A]</h3>

b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:

Rate = k×[A]

0.320M/s = k×[1.50M]

<h3>k = 0.213s⁻¹</h3>

6 0
3 years ago
Suppose you were preparing 1.0 L of a bleaching solution in a volumetric flask, and it calls for 0.21 mol of NaOCl. If all you h
yulyashka [42]

Answer:

0.256 L  

Explanation:

We should use the following formula:

concentration (1) × volume (1) =  concentration (2) × volume (2)

concentration (1) = 0.82 M NaOCl

volume (1) = ?

concentration (2) = 0.21 M NaOCl

volume (2) = 1 L

volume (1) = [concentration (2) × volume (2)] / concentration (1)

volume (1) = [0.21 / 1] / 0.82 = 0.256 L

3 0
3 years ago
Read 2 more answers
How many grams of sodium are in<br> 1.000 mole of sodium?
jolli1 [7]

Answer:

Explanation:

How many atoms are in a 3.5 g sample of sodium (Na)? In this example, multiply the grams of Na by the conversion factor 1 mol Na/ 22.98 g Na, with 22.98g being the molar mass of one mole of Na, which then allows cancelation of grams, leaving moles of Na.

8 0
3 years ago
Read 2 more answers
When modeling a scientific process, it is more important to organize the parts in a way that makes sense to you than to list the
AnnyKZ [126]

The answer would actually be false. I just took the test.

7 0
3 years ago
Read 2 more answers
The activation energy of an uncatalyzed reaction is 70 kJ/mol. When a catalyst is added, the activation energy (at 20 °C) is 42
denis23 [38]

Answer:

T = 215.33 °C

Explanation:

The activation energy is given by the Arrhenius equation:

k = Ae^{\frac{-Ea}{RT}}

<u>Where:</u>

k: is the rate constant

A: is the frequency factor    

Ea: is the activation energy

R: is the gas constant = 8.314 J/(K*mol)

T: is the temperature

We have for the uncatalyzed reaction:

Ea₁ = 70 kJ/mol

And for the catalyzed reaction:

Ea₂ = 42 kJ/mol

T₂ = 20 °C = 293 K

The frequency factor A is constant and the initial concentrations are the same.

Since the rate of the uncatalyzed reaction (k₁) is equal to the rate of the catalyzed reaction (k₂), we have:

k_{1} = k_{2}

Ae^{\frac{-Ea_{1}}{RT_{1}}} = Ae^{\frac{-Ea_{2}}{RT_{2}}}   (1)

By solving equation (1) for T₁ we have:

T_{1} = \frac{T_{2}*Ea_{1}}{Ea_{2}} = \frac{293 K*70 kJ/mol}{42 kJ/mol} = 488. 33 K = 215.33 ^\circ C  

Therefore, we need to heat the solution at 215.33 °C so that the rate of the uncatalyzed reaction is equal to the rate of the catalyzed reaction.

I hope it helps you!      

4 0
4 years ago
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