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Lelu [443]
3 years ago
7

The rate constant for a particular zero-order reaction is 0.075 M s-1. If the initial concentration of reactant is 0.537 M it ta

kes ________ s for the concentration to decrease to 0.100 M.
Chemistry
1 answer:
Anarel [89]3 years ago
8 0

Answer:

It takes 5.83s to decrease the concentration of the reactant from 0.537M to 0.100M

Explanation:

A zero-order reaction follows the equation:

[A] = [A]₀ - kt

<em>Where [A] is actual reaction of the reactant = 0.100M</em>

<em>[A]₀ the initial concentration = 0.537M</em>

<em>k is rate constant = 0.075Ms⁻¹</em>

<em>And t is time it takes:</em>

<em />

0.100M = 0.537M -0.075Ms⁻¹t

-0.437M = -0.075Ms⁻¹t

5.83s = t

It takes 5.83s to decrease the concentration of the reactant from 0.537M to 0.100M

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<u>Answer:</u> The partial pressure of hydrogen is 93.9 kPa.

<u>Explanation:</u>

To calculate the partial pressure of hydrogen, we will follow Dalton's Law.

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Mathematically,

p_{total}=p_A+p_B

According to the question,

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We are given:

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Putting values in above equation, we get:

97.1kPa=3.2kPa+p_{H_2}

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Upon combustion, a 1.3109 g sample of a compound containing only carbon, hydrogen, and oxygen produces 3.2007 g<img src="https:/
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Answer:

The answer to your question is:   C₃H₃O  This is my answer.

Explanation:

Data

Sample = 1.3109 g

CxHyOz

CO₂ = 3.2007 g

H₂O = 1.3102 g

Empirical formula = ?

MW CO2 = 44 g

MW H2O = 18 g

For Carbon

                                     44 g -------------------- 12 g

                                     3.2007 g ------------    x

                                      x = (3.2007 x 12) / 44

                                      x = 0.8729 g of Carbon

                                     12 g of C --------------  1 mol

                                     0.8729 g --------------  x

                                     x = (0.8729 x 1) / 12

                                     x = 0.0727 mol of Carbon

For Hydrogen

                                 18 g ---------------------- 1 g

                              1.3102 g -------------------  x

                                   x = (1.3102 x 1) / 18

                                  x = 0.0727 g of Hydrogen                                      

                                  1 g ------------------------ 1 mol

                                  0.0727g ----------------  x

                                  x = (0.0727 x 1)/1

                                  x = 0.0727 mol of Hydrogen

For oxygen

                    g of Oxygen = g of sample - g of Carbon - g of hydrogen

                    g of Oxygen = 1.3109 - 0.8709 - 0.0727

                    g of Oxygen = 0.3673

                                  16 g of Oxygen ------------- 1 mol of O

                                  0.3673 g ---------------------   x

                                   x = (0.3673 x 1)/ 16

                                   x = 0.0230 mol of Oxygen

Divide by the lowest number of moles

Carbon              0.0727 / 0.023  = 3.1  ≈ 3

Hydrogen         0.0727 / 0.023 = 3.1  ≈ 3

Oxygen             0.0230 / 0.023 = 1

                                        C₃H₃O

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