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Murljashka [212]
3 years ago
9

If the change of enthalpy of this reaction when proceeding left to right is +14 kcal, which chemical equation is correct?

Chemistry
1 answer:
bagirrra123 [75]3 years ago
7 0

Answer:

N₂O₄ + 14 kcal ⇄ 2NO₂.

Explanation:

Since the sign of ΔH determines either the reaction is exothermic or endothermic:

+ve, the reaction is endothermic.

-ve, the reaction is exothermic.

∵ The change of enthalpy of this reaction when proceeding left to right is + 14 kcal (+ ve sign).

∴ The reaction is endothermic, the heat is a part of the reacatnts in the reaction.

So, the reaction is:

N₂O₄ + 14 kcal ⇄ 2NO₂.

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Answer:

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Explanation:

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3 years ago
Balance the equation by inserting coefficients as needed. equation: ZnS + HBr -> ZnBr_{2} + H_{2}S ZnS + HBr ⟶ ZnBr 2 + H 2 S
yawa3891 [41]

Answer : The balanced chemical reaction will be,

ZnS+2HBr\rightarrow ZnBr_2+H_2S

Explanation :

Balanced chemical reaction : It is defined as the reaction in which the number of atoms of individual elements present on reactant side must be equal to the product side.

If the amount of atoms of each type on the left and right sides of a reaction differs then to balance the equation by adding coefficient in the front of the elements or molecule or compound in the chemical equation.

The coefficient tell us about that how many molecules or atoms present in the chemical equation.

The given chemical reaction is,

ZnS+HBr\rightarrow ZnBr_2+H_2S

This reaction is an unbalanced chemical reaction because in this reaction number of hydrogen bromine atoms are not balanced.

In order to balance the chemical equation, the coefficient '2' put before the HBr and we get the balanced chemical equation.

The balanced chemical reaction will be,

ZnS+2HBr\rightarrow ZnBr_2+H_2S

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3 years ago
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Which spectroscopic tool would be best for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane?
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1-H NMR spectroscopy tool will be used for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane.

The preferred method for determining or validating the structure of organic molecules or those containing protons is H NMR. When compared to other nuclei, a solution-state proton spectrum may be obtained relatively quickly, and it contains a wealth of knowledge regarding a compound's structure.

It can be calculated by simply counting the number of unique hydrogens on one side of the symmetry plane will give you the count of signals individual molecules emit in a 1H NMR spectrum.

Therefore, 1-H NMR spectroscopy tool will be used for distinguishing a sample of 1,2,2-tribromopropane from 1,1,2-tribromopropane.

To know more about 1-H NMR spectroscopy

brainly.com/question/20111886

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5 0
2 years ago
A sample of an unknown metal has a mass of 58.932g. it has been heated to 101.00 degrees C, then dropped quickly into 45.20 mL o
yaroslaw [1]
<h3>Answer:</h3>

0.111 J/g°C

<h3>Explanation:</h3>

We are given;

  • Mass of the unknown metal sample as 58.932 g
  • Initial temperature of the metal sample as 101°C
  • Final temperature of metal is 23.68 °C
  • Volume of pure water = 45.2 mL

But, density of pure water = 1 g/mL

  • Therefore; mass of pure water is 45.2 g
  • Initial temperature of water = 21°C
  • Final temperature of water is 23.68 °C
  • Specific heat capacity of water = 4.184 J/g°C

We are required to determine the specific heat of the metal;

<h3>Step 1: Calculate the amount of heat gained by pure water</h3>

Q = m × c × ΔT

For water, ΔT = 23.68 °C - 21° C

                       = 2.68 °C

Thus;

Q = 45.2 g × 4.184 J/g°C × 2.68°C

    = 506.833 Joules

<h3>Step 2: Heat released by the unknown metal sample</h3>

We know that, Q =  m × c × ΔT

For the unknown metal, ΔT = 101° C - 23.68 °C

                                              = 77.32°C

Assuming the specific heat capacity of the unknown metal is c

Then;

Q = 58.932 g × c × 77.32°C

   = 4556.62c Joules

<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
  • We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
  • Therefore;

4556.62c Joules = 506.833 Joules

c = 506.833 ÷4556.62

  = 0.111 J/g°C

Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C

8 0
3 years ago
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