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Galina-37 [17]
3 years ago
10

PLEASE HELP!!! I will mark Brainliest!!! How could you know if two substances were the same based upon 2 chemical reactions?

Chemistry
2 answers:
vladimir1956 [14]3 years ago
8 0

Answer:

Observing what the substance looks (color, size, texture) or smells like (odor). Objects can be made up of the same substance, but they differ in the properties of size, mass, and volume.

Explanation:

The five basic types of chemical reactions are combination, decomposition, single-replacement, double-replacement, and combustion. Analyzing the reactants and products of a given reaction will allow you to place it into one of these categories. Some reactions will fit into more than one category.

Novay_Z [31]3 years ago
5 0

Answer:

1.Color Change.

2.Production of an odor.

3.Change of Temperature.

4.Evolution of a gas (formation of bubbles)

5.Precipitate (formation of a solid)

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Is powdered drink and water a mixture or a solution
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It is a homogenous solution because the powder is suspended in the liquid and there is no difference between any parts of it.
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Help please <br> What percentage of the world does not have a reliable source for clean water?
disa [49]

Answer:9%

Explanation:

7 0
3 years ago
Rank the following solutions in order of increasing acidity (least acidic to most acidic): I: [H3O+] = 1x10−5 II: [OH−] = 1x10−1
kozerog [31]

Answer: option C) II < III < I

i.e [OH−] < [H3O+] < I

Explanation:

First, obtain the pH value of I and II, then compare both with III.

For I

Recall that pH = -log (H+)

So pH3O = -log (H3O+)

= - log (1x10−5)

= 4

For II

pOH = - log(OH-)

= - log(1x10−10)

= 9

For III

pH = 6

Since, pH range from 1 to 14, with values below 7 to be acidic, 7 to be neutral, above 7 to be alkaline: then, 9 < 6 < 4

Thus, the following solutions from least acidic to most acidic is II < III < I

6 0
3 years ago
Use bond energies to calculate the enthalpy of reaction for the combustion of ethane. Average bond energies in kJ/mol C-C 347, C
ivanzaharov [21]

The enthalpy of reaction for the combustion of ethane 2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O calculated from the average bond energies of the compounds is -2860 kJ/mol.

The reaction is:

2CH₃CH₃ + 7O₂ → 4CO₂ + 6H₂O  (1)  

The enthalpy of reaction (1) is given by:

\Delta H = \Delta H_{r} - \Delta H_{p}   (2)

Where:

r: is for reactants

p: is for products

The bonds of the compounds of reaction (1) are:

  • 2CH₃CH₃: 2 moles of 6 C-H bonds + 2 moles of 1 C-C bond
  • 7O₂: 7 moles of 1 O=O bond  
  • 4CO₂: 4 moles of 2 C=O bonds  
  • 6H₂O: 6 moles of 2 H-O bonds

Hence, the enthalpy of reaction (1) is (eq 2):

\Delta H = \Delta H_{r} - \Delta H_{p}

\Delta H = 2*\Delta H_{CH_{3}CH_{3}} + 7\Delta H_{O_{2}} - (4*\Delta H_{CO_{2}} + 6*\Delta H_{H_{2}O})      

\Delta H = 2*(6*\Delta H_{C-H} + \Delta H_{C-C}) + 7\Delta H_{O=O} - (4*2*\Delta H_{C=O} + 6*2*\Delta H_{H-O})  

\Delta H = [2*(6*413 + 347) + 7*498 - (4*2*799 + 6*2*467)] kJ/mol  

\Delta H = -2860 kJ/mol          

Therefore, the enthalpy of reaction for the combustion of ethane is -2860 kJ/mol.

Read more here:

brainly.com/question/11753370?referrer=searchResults  

I hope it helps you!        

7 0
3 years ago
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