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vekshin1
3 years ago
8

What is the volume of 2.268mol of oxygen gas at 35○C and 3.0atm?

Chemistry
1 answer:
Eduardwww [97]3 years ago
6 0
The volume is 

3 x_{123}
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A strong acid like HCl donates its proton so readily that there is essentially no tendency for the conjugate base Cl– to reaccept a proton. Consequently, Cl– is a very weak base.
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The bond between two eatoms of the same element in which the electrons are equally shared is known as a nonpolar covalent bond.
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Please with this problem
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<h3>Answer:</h3>

            Empirical Formula  =  C₃H₈O₃

            Molecular Formula  =  C₃H₈O₃

<h3>Solution:</h3>

Data Given:

                      Mass of Sample  =  9.2 g

                      Mass of Carbon  =  3.6 g

                      Mass of Hydrogen  =  0.8 g

                      Mass of Oxygen  =  9.2 - (3.6 + 0.8) = 4.8 g

Step 1: Calculate Moles of each Element;

                      Moles of C  =  Mass of C ÷ At.Mass of C

                      Moles of C  = 3.6 ÷ 12.01

                      Moles of C  =  0.2997 mol


                      Moles of H  =  Mass of H ÷ At.Mass of H

                      Moles of H  = 0.8 ÷ 1.01

                      Moles of H  =  0.7920 mol


                      Moles of O  =  Mass of O ÷ At.Mass of O

                      Moles of O  = 4.8 ÷ 16.0

                      Moles of O  =  0.3000 mol

Step 2: Find out mole ratio and simplify it;

                C                                        H                                     O

            0.2997                              0.7920                           0.3000

    0.2997/0.2997                  0.7920/0.2997              0.3000/0.2997

               1                                      2.64                                    1.001

Multiply by 3,

               3                               7.92 ≈ 8                                    3

Hence,  Empirical Formula  =  C₃H₈O₃

Step 3: Calculating Molecular Formula:

Molecular formula is calculated by using following formula,

                    Molecular Formula  =  n × Empirical Formula  ---- (1)

Also, n is given as,

                     n  =  Molecular Weight / Empirical Formula Weight

Molecular Weight  =  92 g.mol⁻¹

Empirical Formula Weight  =  12 (C₃) + 1.01 (H₈) + 16 (O₃)  =  92.08 g.mol⁻¹

So,

                     n  =  92 g.mol⁻¹ ÷ 92 g.mol⁻¹

                     n  =  1

Putting Empirical Formula and value of "n" in equation 1,

                    Molecular Formula  = 1 × C₃H₈O₃

                    Molecular Formula  =  C₃H₈O₃

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