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Nana76 [90]
3 years ago
13

At a certain temperature this reaction follows second-order kinetics with a rate constant of 0.00317sâ1: 2N2O5(g) â2N2O4(g) + O2

9(g) Suppose a vessel contains SO3 at a concentration of 1.44M . Calculate the concentration of SO3 in the vessel 0.240 seconds later. You may assume no other reaction is important.Round your answer to 2 significant digits.
Chemistry
1 answer:
Triss [41]3 years ago
8 0

Answer:

[A] = 1.438M = 1.4M (Two s.f)

Explanation:

Rate constant, k =  0.00317

Initial Concentration, [A]o = 1.44M

Final Concentration, [A] = ?

Time, t = 0.240 s

Since this is a second order reaction, the formula for this is given as;

1 / [A] = 1 / [A]o + kt

1 / [A] = 1 / 1.44 + (0.00317 * 0.240)

1 / [A] = 0.6944 + 0.0007608

1 / [A] = 0.6952

[A] = 1.438M = 1.4M (Two s.f)

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How many grams of sodium sulfate Na2SO4 (mw = 142.04 g/mol) is needed to prepare 350.0 ml of a solution having a concentration o
ch4aika [34]

According to the definition of molarity, 350 mL of a solution with a sodium ion concentration of 0.125 M requires 6.2125 g of Na2SO4 to manufacture.

<h3><u>How to find Molarity ?</u></h3>

The number of moles of a solute that are dissolved in a given volume is what is meant by the definition of molarity, which is a measurement of a solute's concentration.

By dividing the moles of the solute by the volume of the solution, molarity, also known as the molar concentration of a solution, is obtained.

Molarity = No. of moles of solute / Volume

Molarity is expressed in units mole/litre

In this case, you know that:

  • molarity= 0.125 M
  • number of moles of solute= ?
  • volume= 350 mL= 0.350 mL

Replacing in the definition of molarity:

0.125M = No. of moles of solute / 0.350l

Solving:

number of moles of solute= 0.125 M× 0.350 L

number of moles of solute= 0.04375 moles

Being 142 g/mole the molar mass of Na₂SO₄, that is, the mass of one mole of the compound, the amount of mass that contains 0.04375 moles is calculated as:

mass= 0.04375 moles× 142 g/moles

mass= 6.2125 g

In summary, 6.2125 g of Na₂SO₄ is needed to prepare 350 mL of a solution having a sodium ion concentration of  0.125 M.

To view more about concentrations, refer to;

brainly.com/question/10725862

#SPJ4

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