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KonstantinChe [14]
3 years ago
14

What is the molarity of a solution that contains 2.14 moles (CH3)2SO in 2.00 L solution?

Chemistry
1 answer:
seraphim [82]3 years ago
3 0
<h3>Answer:</h3>

1.07 M

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

  • Molarity = moles of solute / liters of solution
<h3>Explanation:</h3>

<u>Step 1: Define</u>

[Given] 2.14 moles (CH₃)₂SO

[Given] 2.00 L

[Solve] Molarity

<u>Step 2: Solve</u>

  1. Substitute in variables [Molarity]:                                                                    x M = 2.14 moles (CH₃)₂SO / 2.00 L solution
  2. [Molarity] Divide:                                                                                               x = 1.07 M
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Why do nitrogen and oxygen form negative ions, not positive ones, in simple binary compounds.
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2 years ago
A reaction was performed in which 4.0 g of cyclohexanol was reacted with an acid catalyst to obtain 2.8 g of cyclohexene. calcul
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Answer:

              %age Yield  =  85.36 %

Solution:

The Balance Chemical Reaction is as follow,

               C₆H₁₂O  +   Acid Catalyst    →    C₆H₁₀  +  Acid Catalyst + H₂O

According to Equation ,

              100 g (1 mole) C₆H₁₂O produces  =  82 g (1 moles) of C₆H₁₀

So,

                       4.0 g of C₆H₁₂O will produce  =  X g of C₆H₁₀

Solving for X,

                      X =  (4.0 g × 82 g) ÷ 100 g

                      X  =  3.28 g of C₆H₁₀   (Theoretical Yield)

As we know,

                     %age Yield  =  (Actual Yield ÷ Theoretical Yield) × 100

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5 0
3 years ago
Under certain conditions the rate of this reaction is zero order in dinitrogen monoxide with a rate constant of 0.0067·Ms−1: 2N2
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Answer:

5.6 seconds

Explanation:

The reaction follows a zero-order in dinitrogen monoxide

Rate = k[N20]^0 = change in concentration/time

[N20]^0 = 1

Time = change in concentration of N2O/k

Initial number of moles of N2O = 300 mmol = 300/1000 = 0.3 mol

Initial concentration = moles/volume = 0.3/4 = 0.075

Number of moles after t seconds = 150 mmol = 150/1000 = 0.15 mol

Concentration after t seconds = 0.15/4 = 0.0375 M

Change in concentration of N2O = 0.075 - 0.0375 = 0.0375 M

k = 0.0067 M/s

Time = 0.0375/0.0067 = 5.6 s

4 0
3 years ago
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