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Black_prince [1.1K]
2 years ago
3

You walk in the front door of your house. you smell an onion that someone is cutting in the kitchen. explain why you are able to

smell it, what this process is called, and what other types of substances exhibit this behavior.
Chemistry
1 answer:
Y_Kistochka [10]2 years ago
4 0
Plants let out volatile organic compounds as a chemical defense and first aid. this can be found when the plant is being injured; like cutting grass and chopping an onion.
You might be interested in
NH4NO3, whose heat of solution is 25.7 kJ/mol, is one substance that can be used in cold pack. If the goal is to decrease the te
makvit [3.9K]

Answer:

There are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

Explanation:

To decrease the temperature of the solution there are necessaries:

4,184J/g°C×(5,0°C-25,0°C)×(100,0g+X) = -Y

8368J + 83,68J/gX = Y <em>(1)</em>

Where x are grams of NH₄NO₃ you need to add and Y is the energy that you need to decrease the heat.

Also, the energy Y will be:

Y = 25700J/mol×\frac{1mol}{80,043g}X

Y = 321J/g X <em>(2)</em>

Replacing (2) in (1)

8368J + 83,68J/g X = 321J/g X

8363J = 237,32J/gX

<em>X = 35,2g</em>

<em />

Thus, there are necessaries 35,2g of NH₄NO₃ per 100,0g of water to decrease the temperature of the solution from 25,0°C to 5,0°C

I hope it helps!

6 0
2 years ago
Which of the following compounds has polar covalent bonds: NaBr, Br2, HBr, and CBr4?
svetlana [45]

Answer: Option (e) is the correct answer.

Explanation:

A bond that is formed when an electron is transferred from one atom to another results in the formation of an ionic bond.

For example, NaBr will be an ionic compound as there is transfer of electron from Na to Br.

Whereas a bond that is formed by sharing of electrons is known as a covalent bond.

For example, CBr_{4} will be a covalent compound as there is sharing of electron between carbon and bromine atom.

Also, when electrons are shared between the combining atoms and there is large difference in electronegativity of these atoms then partial charges develop on these atoms. As a result, it forms a polar covalent bond.

For example, in a HBr compound there is sharing of electrons between H and Br. Also, due to difference in electronegativity there will be partial positive charge on H and partial negative charge on Br.  

Thus, we can conclude that out of the given options HBr is the only compound that has polar covalent bonds.

8 0
2 years ago
What is the net cell reaction for the iron-silver voltaic cell?
RSB [31]
The net cell reaction for the iron-silver voltaic cell would be:

<span>3Ag+ + Fe --------> 3Ag + Fe3+
</span>
wherein it comes from the reaction of the cathode and the anode:

<span>Ag+ + e- -------> Ag, for the cathode and;
</span>
<span>Fe -------> Fe3+ + 3e-, for the anode
</span>
When both equations would be multiplied, the it would now yield the balanced net reaction as stated above.
4 0
2 years ago
How many inner, outer, and valence electrons are present in an atom of each of the following elements?
Aneli [31]

<u>Answer:</u>

<u>For a:</u> The number of inner electrons, outer electrons and valence electrons are 28, 7 and 7 respectively.

<u>For b:</u> The number of inner electrons, outer electrons and valence electrons are 54, 1 and 1 respectively.

<u>For c:</u> The number of inner electrons, outer electrons and valence electrons are 23, 1 and 1 respectively.

<u>For d:</u> The number of inner electrons, outer electrons and valence electrons are 36, 2 and 2 respectively.

<u>For e:</u> The number of inner electrons, outer electrons and valence electrons are 2, 7 and 7 respectively.

<u>Explanation:</u>

Outer shell electrons are the electrons which are not tightly held by the electrons. They are called as valence electrons. The electrons present in the highest principle quantum number are known as valence electrons.

Inner shell electrons are the electrons which are tightly held by the electrons. They are called as core electrons.

Inner electrons = Total number of electrons - Valence electrons

Total number of electrons in an atom is equal to the atomic number of the element.

For the given options:

  • <u>Option a:</u>  Bromine (Br)

Bromine is the 35th element of the periodic table having electronic configuration of 1s^22s^22p^63s^23p^64s^23d^{10}4p^5

Highest principle quantum number is 4

Number of valence electrons = 7

Number of outer electrons = 7

Number of inner electrons = 35 - 7 = 28

  • <u>Option b:</u>  Cesium (Cs)

Cesium is the 55th element of the periodic table having electronic configuration of 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^24d^{10}5p^66s^1

Highest principle quantum number is 6

Number of valence electrons = 1

Number of outer electrons = 1

Number of inner electrons = 55 - 1 = 54

  • <u>Option c:</u>  Chromium (Cr)

Chromium is the 24th element of the periodic table having electronic configuration of 1s^22s^22p^63s^23p^64s^13d^{5}

Highest principle quantum number is 4

Number of valence electrons = 1

Number of outer electrons = 1

Number of inner electrons = 24 - 1 = 23

  • <u>Option d:</u>  Strontium (Sr)

Strontium is the 38th element of the periodic table having electronic configuration of 1s^22s^22p^63s^23p^64s^23d^{10}4p^65s^2

Highest principle quantum number is 5

Number of valence electrons = 2

Number of outer electrons = 2

Number of inner electrons = 38 - 2 = 36

  • <u>Option e:</u>  Fluorine (F)

Fluorine is the 9th element of the periodic table having electronic configuration of 1s^22s^22p^5

Highest principle quantum number is 2

Number of valence electrons = 7

Number of outer electrons = 7

Number of inner electrons = 9 - 7 = 2

4 0
2 years ago
A student reacts 20.0 mL of 0.248 M H2SO4 with 15.0 mL of 0.195 M NaOH. Write a balanced chemical equation to show this reaction
Molodets [167]

<u>Answer:</u> The concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

<u>Explanation:</u>

To calculate the number of moles for given molarity, we use the equation:

\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}     .....(1)

<u>For NaOH:</u>

Initial molarity of NaOH solution = 0.195 M

Volume of solution = 15.0 mL = 0.015 L   (Conversion factor:   1 L = 1000 mL)

Putting values in equation 1, we get:

0.195M=\frac{\text{Moles of NaOH}}{0.015L}\\\\\text{Moles of NaOH}=(0.195mol/L\times 0.015L)=2.925\times 10^{-3}mol

<u>For sulfuric acid:</u>

Initial molarity of sulfuric acid solution = 0.248 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

0.248M=\frac{\text{Moles of }H_2SO_4}{0.020L}\\\\\text{Moles of }H_2SO_4=(0.248mol/L\times 0.020L)=4.96\times 10^{-3}mol

The chemical equation for the reaction of NaOH and sulfuric acid follows:

2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

By Stoichiometry of the reaction:

2 moles of NaOH reacts with 1 mole of sulfuric acid

So, 2.925\times 10^{-3} moles of KOH will react with = \frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol of sulfuric acid

As, given amount of sulfuric acid is more than the required amount. So, it is considered as an excess reagent.

Thus, NaOH is considered as a limiting reagent because it limits the formation of product.

Excess moles of sulfuric acid = (4.96-1.462)\times 10^{-3}=3.498\times 10^{-3}mol

By Stoichiometry of the reaction:

2 moles of KOH produces 1 mole of sodium sulfate

So, 2.925\times 10^{-3} moles of KOH will produce = \frac{1}{2}\times 2.925\times 10^{-3}=1.462\times 10^{-3}mol of sodium sulfate

  • <u>For sodium sulfate:</u>

Moles of sodium sulfate = 1.462\times 10^{-3}moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

\text{Molarity of sodium sulfate}=\frac{1.462\times 10^{-3}}{0.035}=0.0418M

  • <u>For sulfuric acid:</u>

Moles of excess sulfuric acid = 3.498\times 10^{-3}mol

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

\text{Molarity of sulfuric acid}=\frac{3.498\times 10^{-3}}{0.035}=0.0999M

  • <u>For NaOH:</u>

Moles of NaOH remained = 0 moles

Volume of solution = [15.0 + 20.0] = 35.0 mL = 0.035 L

Putting values in equation 1, we get:

\text{Molarity of NaOH}=\frac{0}{0.050}=0M

Hence, the concentration of salt (sodium sulfate), sulfuric acid and NaOH in the solution is 0.0418 M, 0.0999 M and 0 M respectively.

8 0
2 years ago
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