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g100num [7]
2 years ago
10

If you have 12 atoms of hydrogen before a chemical reaction how many atoms of hydrogen will be president after the chemical reac

tion
Physics
1 answer:
kaheart [24]2 years ago
4 0
If you have 12 atoms of hydrogen before a chemical reaction, the number of hydrogen atoms that will be present after the chemical reaction is 12 atoms.
The Law of Conservation of Mass (LOCOM) states that mass is neither created nor destroyed before and after any chemical reaction.
According to the Law of Conservation of Mass (LOCOM), a balanced chemical equation requires that the number of atoms on the reactant side must be equal to the number of atoms on the product side of any chemical reaction.
In this context, a chemical reaction having 12 atoms of hydrogen as reactants at the beginning, should also produce a total of 12 atoms of hydrogen as products at the end of the chemical reaction.
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Solve for work when
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So, <u>the value of the work is approximately 84.65 J</u>.

<h2>Introduction</h2>

Hi ! Here I will help you to discuss the subject about work that caused by force in amount value of angle. Work is affected by the force and displacement.

  • If related to the magnitude of the force, the amount of work will be proportional to the magnitude of the applied force. Thats mean, if the value of the force that applied on it is greater, then the value of the work will be greater.
  • If related to the magnitude of shift, the amount of work will be proportional to the magnitude of shift of object. Thats mean, if the value of the shift on it is greater, then the value of the work will be greater.
<h3>Formula Used</h3>

The work done by a moving object can be expressed in the equation:

If the Angle Is Ignored

\boxed{\sf{\bold{W = F \times s}}}

If the Angle Effect on Work

\boxed{\sf{\bold{W = F \times s \times \cos(\theta)}}}

With the following condition:

  • W = work that done by object (J)
  • F = force that applied (N)
  • s = shift or distance (m)
  • \sf{\theta} = angle of elevation (°)

<h3>Solution</h3>

We know that :

  • F = force that applied = \sf{1.41 \times 10^4} N
  • s = shift or distance = 84.9 m
  • \sf{\theta} = angle of elevation = 45°

What was asked ?

  • W = work that done by object = ... J

Step by step :

\sf{W = F \times s \times \cos(\theta)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \cos(45^o)}

\sf{W = (1.41 \cdot 10^4) \times 84.9 \times \frac{\sqrt{2}}{2}}

\sf{W = 119.709 \times \frac{\sqrt{2}}{2}}

\sf{W = 59.8545 \sqrt{2}}

\boxed{\sf{W \approx 84.65 \: J}}

<h3>Conclusion</h3>

So, the value of the work is approximately 84.65 J.

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