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(1) The distance between a carbon atom \((m=12 \mathrm{u})\) and an oxygen atom \((m=16 \mathrm{u})\) in the \(\mathrm{CO}\) molecule is \(1.13 \times 10^{-10} \mathrm{m} .\) How far from the carbon atom is the center of mass of the molecule?

Short Answer

Expert verified
The center of mass is about \(6.457 \times 10^{-11} \mathrm{m}\) from the carbon atom.

Step by step solution

01

Understand the Problem

We need to find the center of mass of a molecule, specifically the distance from the carbon atom (mass 12 u) to the center of mass of the CO molecule, given the distance between the carbon and oxygen atoms.
02

Identify the Symbols and Units

Here, the mass of the carbon atom is given as \(m_1 = 12 \mathrm{u}\) and the mass of the oxygen atom as \(m_2 = 16 \mathrm{u}\). The distance between the two atoms is \(d = 1.13 \times 10^{-10} \mathrm{m}\). The center of mass \(x_{cm}\) can be determined using these values.
03

Write the Formula for Center of Mass

The formula to calculate the center of mass \(x_{cm}\) for a two-point mass system along a line is:\[x_{cm} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2}\]where \(x_1\) and \(x_2\) are the positions of mass \(m_1\) and \(m_2\) respectively. Without loss of generality, let's place the carbon atom at position \(x_1 = 0\) and the oxygen atom at \(x_2 = d\).
04

Substitute the Values into the Formula

Substituting the known values:- \(m_1 = 12\)- \(m_2 = 16\)- \(x_1 = 0\)- \(x_2 = 1.13 \times 10^{-10} \mathrm{m}\)we have:\[x_{cm} = \frac{12 \cdot 0 + 16 \cdot 1.13 \times 10^{-10}}{12 + 16}\]
05

Calculate the Center of Mass

Calculate the numerator:\[16 \times 1.13 = 18.08\]Thus the equation becomes:\[x_{cm} = \frac{18.08 \times 10^{-10}}{28}\]Calculate the denominator and simplify:\[x_{cm} = 0.6457 \times 10^{-10} \mathrm{m} = 6.457 \times 10^{-11} \mathrm{m}\]
06

Conclusion

The center of mass of the CO molecule is located approximately \(6.457 \times 10^{-11} \mathrm{m}\) from the carbon atom.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

CO Molecule
The CO molecule, or carbon monoxide molecule, is a simple yet fascinating example in molecular physics and chemistry. This molecule consists of only two different types of atoms: one carbon atom and one oxygen atom. Understanding the structure and properties of such a diatomic molecule can help us learn about more complex molecules.
One core concept when studying a CO molecule is the idea of the center of mass. In molecules, the center of mass is similar to a balance point, where the mass of different atoms is evenly distributed around it. This point doesn't have to be at the geometric center due to the differing atomic masses.
In a CO molecule, since the atomic mass of oxygen is greater than that of carbon, the center of mass is closer to the oxygen atom. This distribution can affect how the molecule interacts with others, and understanding it is key in fields like spectroscopy and quantum chemistry.
Mass Calculation
Mass calculation in a molecule involves identifying the individual masses of the constituent atoms and understanding how these masses contribute to the properties of the molecule.
In our example, we consider the masses of a carbon atom and an oxygen atom. With atomic masses of 12 u for carbon and 16 u for oxygen, these individual masses play an integral role in determining the center of mass of the molecule. The unit 'u' stands for atomic mass units, a standard unit used to express atomic and molecular weights.
When calculating the center of mass, these masses are used alongside their respective positions. The formula \[x_{cm} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2}\]is essential as it helps in analyzing how the masses are spatially arranged along with their positions. By knowing the atomic masses and their spatial distribution, one can understand more about the molecule's behavior in interactions and reactions.
Distance Between Atoms
The distance between atoms in a molecule is a fundamental aspect when studying their structure. This distance tells us how far apart the nuclei of the atoms are from each other. For a CO molecule, this distance is given as \(1.13 \times 10^{-10}\) meters.
Knowing this distance is crucial because it helps in calculating the center of mass and understanding molecular geometry. By placing the coordinates of the carbon and oxygen atoms along a line, where we often set the carbon atom at the origin (0 position), we can use the distance to position the oxygen atom appropriately in our calculations.
This knowledge aids not just in theoretical calculations, but also in practical applications like molecular design and analysis. Understanding interatomic distances can also help predict molecular properties such as vibrational modes and even chemical reactivity in more complex processes.

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Most popular questions from this chapter

(II) A neutron collides elastically with a helium nucleus (at rest initially) whose mass is four times that of the neutron. The helium nucleus is observed to move off at an angle \(\theta_{\mathrm{He}}^{\prime}=45^{\circ} .\) Determine the angle of the neutron, \(\theta_{\mathrm{n}}^{\prime},\) and the speeds of the two particles, \(v_{\mathrm{n}}^{\prime}\) and \(v_{\mathrm{He}}^{\prime},\) after the collision. The neutron's initial speed is \(6.2 \times 10^{5} \mathrm{~m} / \mathrm{s}\).

(III) A huge balloon and its gondola, of mass \(M,\) are in the air and stationary with respect to the ground. A passenger, of mass \(m\), then climbs out and slides down a rope with speed \(v\), measured with respect to the balloon. With what speed and direction (relative to Earth) does the balloon then move? What happens if the passenger stops?

(II) Determine the fraction of kinetic energy lost by a neutron \(\left(m_{1}=1.01 \mathrm{u}\right)\) when it collides head-on and elastically with a target particle at rest which is \((a){ }_{1}^{1} \mathrm{H}(m=1.01 \mathrm{u})\) (b) \({ }_{1}^{2} \mathrm{H}\) (heavy hydrogen, \(\left.m=2.01 \mathrm{u}\right) ;(c){ }_{6}^{12} \mathrm{C}(m=12.00 \mathrm{u})\) (d) \({ }^{208} \mathrm{~Pb}\) (lead, \(m=208 \mathrm{u}\) ).

The space shuttle launches an \(850-\) kg satellite by ejecting it from the cargo bay. The ejection mechanism is activated and is in contact with the satellite for 4.0 s to give it a velocity of 0.30 \(\mathrm{m} / \mathrm{s}\) in the \(z\) -direction relative to the shuttle. The mass of the shuttle is \(92,000 \mathrm{kg}\) . (a) Determine the component of velocity \(v_{\mathrm{f}}\) of the shuttle in the minus z-direction resulting from the ejection. \((b)\) Find the average force that the shuttle exerts on the satellite during the ejection.

(III) A scale is adjusted so that when a large, shallow pan is placed on it, it reads zero. A water faucet at height \(h=2.5 \mathrm{~m}\) above is turned on and water falls into the pan at a rate \(R=0.14 \mathrm{~kg} / \mathrm{s} .\) Determine \((a)\) a formula for the scale reading as a function of time \(t\) and \((b)\) the reading for \(t=9.0 \mathrm{~s} .(c)\) Repeat \((a)\) and \((b),\) but replace the shallow pan with a tall, narrow cylindrical container of area \(A=20 \mathrm{~cm}^{2}\) (the level rises in this case).

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