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ad-work [718]
2 years ago
12

The rate constant for a reaction is 4.65 L mol-1s-1. What is the overall order of the reaction? zero

Chemistry
2 answers:
il63 [147K]2 years ago
8 0

Answer:

second order

Explanation:

units of reaction and their order.

Zero order --> M^1 s^-1 = M/s

First order --> M^0 s^-1 = 1/s

Second order --> M^-1 s^-1 = L/mol s

In the question rate constant k =  4.65 L mol-1s-1. = 4.65 L/mol s

Hence, the reaction is a second order reaction

kenny6666 [7]2 years ago
5 0

Answer:

The answer is "second".

Explanation:

Given:

rate constant for a reaction = 4.65 \frac{L}{mol \cdot s}

\text{The rate of unit} =  \frac{m}{s} \ \ \ \  \ \ \ \ \ \ \ \ \ \ \ \  _{where}  \ \ M=\frac{mol}{l}   \\\\

                          =\frac{mol}{l \cdot s}

The orders                                                       unit value of the k

0                                                                               \frac{m}{s}

1                                                                                 \frac{1}{s}

2                                                                               \frac{l}{m s}

The rate of the unit is constant for the second-order that's why it is correct.

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A solution is made by dissolving 23.5 grams of glucose (C6H12O6) in 0.245 kilograms of water. If the molal freezing point consta
aalyn [17]

Answer:

- 0.99 °C ≅ - 1.0 °C.

Explanation:

  • We can solve this problem using the relation:

<em>ΔTf = (Kf)(m),</em>

where, ΔTf is the depression in the freezing point.

Kf is the molal freezing point depression constant of water = -1.86 °C/m,

m is the molality of the solution (m = moles of solute / kg of solvent = (23.5 g / 180.156 g/mol)/(0.245 kg) = 0.53 m.

<em>∴ ΔTf = (Kf)(m)</em> = (-1.86 °C/m)(0.53 m) =<em> - 0.99 °C ≅ - 1.0 °C.</em>

4 0
3 years ago
Is one mole of zinc the same as one atom of zinc?
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Answer: No.

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3 0
2 years ago
If 0.2 g of nitrobenzene are added to 10.9 g of naphthalene, calculate the molality of the solution. (given: molar mass of nitro
In-s [12.5K]

Molality is defined as 1 mole of a solute in 1 kg of solvent.  

Molality=

\frac{Number of moles of solute}{Mass of solvent in kg}

Number of moles of solute, n=  

\frac{Given mass of the substance}{Molar mass of the substance}

Given mass of the nitrobenzene=0.2 g

Molar mass of the substance= 123.06 g mol⁻¹

Number of moles of nitrobenzene,  

n= \frac{0.2 }{123.06}

Number of moles of nitrobenzene, n= 0.0016  mol

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7 0
3 years ago
When 25.0 g of ch4 reacts completely with excess chlorine yielding 45.0 g of ch3cl, what is the percentage yield, according to c
Ber [7]

Taking into account definition of percent yield, the percent yield for the reaction is 57.08%.

<h3>Reaction stoichiometry</h3>

In first place, the balanced reaction is:

CH₄ + Cl₂ → CH₃Cl + HCl

By reaction stoichiometry (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:

  • CH₄: 1 mole
  • Cl₂: 1 mole
  • CH₃Cl: 1  mole
  • HCl:  1 mole

The molar mass of the compounds is:

  • CH₄: 16 g/mole
  • Cl₂: 70.9 g/mole
  • CH₃Cl: 50.45 g/mole
  • HCl:  36.45 g/mole

Then, by reaction stoichiometry, the following mass quantities of each compound participate in the reaction:

  • CH₄: 1 mole ×16 g/mole= 16 grams
  • Cl₂: 1 mole ×70.9 g/mole= 70.9 grams
  • CH₃Cl: 1 mole ×50.45 g/mole= 50.45 grams
  • HCl: 1 mole ×36.45 g/mole= 36.45 grams

Mass of CH₃Cl formed

The following rule of three can be applied: if by reaction stoichiometry 16 grams of CH₄ form 50.45 grams of CH₃Cl, 25 grams of CH₄ form how much mass of CH₃Cl?

mass of CH_{3} Cl=\frac{25 grams of CH_{4}x 50.45grams of CH₃Cl }{16 grams of CH_{4}}

<u><em>mass of CH₃Cl= 78.83 grams</em></u>

Then, 78.83 grams of CH₃Cl can be produced from 25 grams of CH₄.

<h3>Percent yield</h3>

The percent yield is the ratio of the actual return to the theoretical return expressed as a percentage.

The percent yield is calculated as the experimental yield divided by the theoretical yield multiplied by 100%:

percent yield=\frac{actual yield}{theorical yield}x100

where the theoretical yield is the amount of product acquired through the complete conversion of all reagents in the final product, that is, it is the maximum amount of product that could be formed from the given amounts of reagents.

Percent yield for the reaction in this case

In this case, you know:

  • actual yield= 45 grams
  • theorical yield= 78.83 grams

Replacing in the definition of percent yields:

percent yield=\frac{45 grams}{78.83 grams}x100

Solving:

<u><em>percent yield= 57.08%</em></u>

Finally, the percent yield for the reaction is 57.08%.

Learn more about

the reaction stoichiometry:

brainly.com/question/24741074

brainly.com/question/24653699

percent yield:

brainly.com/question/14408642

#SPJ1

5 0
2 years ago
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