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WINSTONCH [101]
2 years ago
10

PLEASE HELP!!!! Draw a representation of John Dalton’s atomic model.

Chemistry
1 answer:
larisa [96]2 years ago
7 0

Answer:

John Dalton's Atomic Model Below ⬇

Explanation:

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Who is the Administrator of Brainly??​<br><br>Hopefully, someone will answer (◍•ᴗ•◍)❤
sleet_krkn [62]

Answer:

Hey mate....

Explanation:

This is ur answer....

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Kavisharma/ Haezel is the admin of brainly......

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3 0
3 years ago
Read 2 more answers
Which one is unsaturated hydrocarbon?
Fittoniya [83]

Answer:

The correct answer is - D C2H4.

Explanation:

Saturated hydrocarbons are hydrocarbons with single covalent C-C bonds. They are known as alkanes. The general formula for these hydrocarbons is  CnH2n+2

Unsaturated hydrocarbons the hydrocarbons with double or triple covalent C-C bonds. They are known as alkenes and alkynes respectively. The general formula for these hydrocarbons is  CnH2n and CnHn-2

For the given options:

Option D: C2H4, is the simplest alkene with a double bond so it is an unsaturated hydrocarbon.

6 0
3 years ago
Why might people add a base to an acidic lake?
spayn [35]

Answer:

To decrease the algae

Explanation:

Acid lakes have more algae than other lakes

6 0
3 years ago
Still using the reaction from the video, what is the limiting reagent when 2 moles of Al are reacted
kondaur [170]

The limiting reagent will be Al

<h3>What are limiting reagents?</h3>

They are reagents that limit the quantity of products that are formed in reactions.

From the equation of the reaction:

4Al + 3O_2 --- > 2Al_2O_3

The mole ratio of Al to O2 is 4:3.

With 2 moles of Al and 2 moles of O2, Al becomes limiting while O2 is in excess.

With 2 moles of O2, the amount of Al required should be:

                          2 x 4/3 = 2.67 moles.

With 2 moles of Al, the amount of O2 required should be:

                    2 x 3/4 = 1.5 moles

Thus, O2 is in excess by 0.5 moles.

More on limiting reagents can be found here: brainly.com/question/11848702

#SPJ1

8 0
2 years ago
The formation of ammonia is represented by the equation N2(g) + 3H2(g) ⇌ 2NH3(g). Determine the enthalpy of formation of ammonia
Savatey [412]

Answer:

\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Explanation:

Hello!

In this case, since the study of the bond energy allows us to compute the enthalpies of some reactions, for this combination reaction by which ammonia is yielded, we understand the enthalpy of reaction equals the enthalpy of formation of ammonia, and, in terms of the bonds energy we can write:

\Delta _fH_{NH_3}=Delta _rH=\Sigma \Delta H(bonds \ broken)-\Sigma \Delta H(bonds \ formed)

Whereas the bonds enthalpy of those bonds that get broken cover the N≡N and the three H-H bonds at the reactants side and the enthalpy of those bonds that are formed cover the six N-H bonds at the products; which means we obtain:

\Delta _fH_{NH_3}=942\frac{kJ}{mol} +3*436\frac{kJ}{mol}-6*386\frac{kJ}{mol}\\\\\Delta _fH_{NH_3}=-66\frac{kJ}{mol}

Which differs from the theoretical value that is -46 kJ/mol.

Best regards!

7 0
3 years ago
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