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Tom [10]
3 years ago
7

What is the instantaneous rate of formation of product C given the following information: a. Stoichiometric equation A+ B2C+ D b

. Applicable rate equation is r.-k"CA"CB c. The rate constant is 6.0 liters/(mole-minute) d. The current concentrations of A and B species are CA 0.6 moles/liter and Ca 0.24 moles/liter

Chemistry
1 answer:
quester [9]3 years ago
6 0

Answer:

1.728 mol /(L*min)

Explanation:

Hello,

In the attached photo, you'll find the numerical procedure for your question.

- Take into account that the negative sign is eligible for reagents and positive for products.

Best regards!

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How many grams of fluorine are contained in 8 molecules of boron trifluoride?
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<h3>Answer:</h3>

             7.57 × 10⁻²² g of F

<h3>Solution:</h3>

Data Given:

                 Number of Molecules  =  8

                 M.Mass of BF₃ =  67.82 g.mol⁻¹

                 Mass of Fluorine atoms  =  ?

Step 1: Calculate Moles of BF₃

           Moles  =  Number of Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

Putting value,

            Moles  =   8 Molecules ÷ 6.022 × 10²³ Molecules.mol⁻¹

            Moles  =  1.33 × 10⁻²³ mol

Step 2: Calculate Mass of BF₃:

                   Moles  =  Mass ÷ M.Mass

Solving for Mass,

                   Mass  =  Moles × M.Mass

Putting values,

                   Mass  =  1.33 × 10⁻²³ mol × 67.82 g.mol⁻¹

                   Mass  =  9.0 × 10⁻²² g

Step 3: Calculate Mass of Fluorine Atoms:

As,

                         67.82 g BF₃ contains  =  57 g of F

So,

                    9.0 × 10⁻²² g will contain  =  X g of F

Solving for X,

                       X =  (9.0 × 10⁻²² g × 57 g) ÷ 67.82 g

                        X  =  7.57 × 10⁻²² g of F

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