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Concept explainers
(a)
To determine:
The reason for bending in the path of the charged particle.
(a)
![Check Mark](/static/check-mark.png)
Answer to Problem 1E
Solution:
The bending of path of the charged particles is because of presence of charge on the plates.
Explanation of Solution
When a charge particle comes in vicinity of another charge it gets either attracted or repelled by the charge. Same charges repel each other and opposite charges attract teach other.
The path of the charged particle bends because when it passes between two electrically charged plates it interacts with the charges on the plate and experiences a force.
The presence of charged plates bends the path of the charged particle.
(b)
To determine:
The sign of electrical charge on the particle.
(b)
![Check Mark](/static/check-mark.png)
Answer to Problem 1E
Solution:
The sign of electrical charge on the particle is negative.
Explanation of Solution
A positive charge gets attracted by a negative charge and vice versa. The particle is bending towards the positively charged plate it means it is getting attracted by the positive charge. So, the charge on the particle must be opposite that is it is negatively charged.
The particle is getting attracted by positively charged plate. So, the charge on it is negative.
(c)
To determine:
The change in bending on increasing the charge on the plate.
(c)
![Check Mark](/static/check-mark.png)
Answer to Problem 1E
Solution:
As the charge on the plates is increased, the bending also increases.
Explanation of Solution
The extent of attraction or repulsion increases on increasing the charge. As the charge on plates is increased the attraction felt by charged particle will increase and hence the bending is also expected to increase.
Increase of charge on plates, increases the bending.
(d)
To determine:
The change in bending on increasing the mass of the particle.
(d)
![Check Mark](/static/check-mark.png)
Answer to Problem 1E
Solution:
As the mass of the particle is increased, the bending decreases.
Explanation of Solution
The extent of attraction or repulsion is inversely proportional to the mass of the particle. As the mass of the particle is increased the attraction felt by charged particle will decrease and hence the bending is also expected to decrease.
Increase of charge on plates, decreases the bending.
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Chapter 2 Solutions
Laboratory Experiments for Chemistry: The Central Science (13th Edition)
- In the solid state, oxalic acid occurs as a dihydrate with the formula H2C2O4 C+2H2O. Use this formula to calculate the formula weight of oxalic acid. Use the calculated formula weight and the number of moles (0.00504mol) of oxalic acid in each titrated unknown sample recorded in Table 6.4 to calculate the number of grams of pure oxalic acid dihydrate contained in each titrated unknown sample.arrow_forward1. Consider a pair of elements with 2p and 4p valence orbitals (e.g., N and Se). Draw their (2p and 4p AO's) radial probability plots, and sketch their angular profiles. Then, consider these orbitals from the two atoms forming a homonuclear л-bond. Which element would have a stronger bond, and why? (4 points)arrow_forwardWrite the reaction and show the mechanism of the reaction. Include the mechanism for formation of the NO2+ 2. Explain, using resonance structures, why the meta isomer is formed. Draw possible resonance structures for ortho, meta and para.arrow_forward
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- 6. a) C2's. Phosphorus pentafluoride PF5 belongs to D3h symmetry group. Draw the structure of the molecule, identify principal axis of rotation and perpendicular (4 points) b) assume that the principal axis of rotation is aligned with z axis, assign symmetry labels (such as a1, b2, etc.) to the following atomic orbitals of the P atom. (character table for this group is included in the Supplemental material). 3s 3pz (6 points) 3dz²arrow_forward2. Construct Lewis-dot structures, and draw VESPR models for the ions listed below. a) SiF5 (4 points) b) IOF4 (4 points)arrow_forward5. Complex anion [AuCl2]¯ belongs to Doh symmetry point group. What is the shape of this ion? (4 points)arrow_forward
- 4. Assign the following molecules to proper point groups: Pyridine N 1,3,5-triazine N Narrow_forward7. a) Under normal conditions (room temperature & atmospheric pressure) potassium assumes bcc lattice. Atomic radius for 12-coordinate K atom is listed as 235 pm. What is the radius of potassium atom under normal conditions? (3 points) b) Titanium metal crystallyzes in hcp lattice. Under proper conditions nitrogen can be absorbed into the lattice of titanium resulting in an alloy of stoichiometry TiNo.2. Is this compound likely to be a substitutional or an interstitial alloy? (Radius of Ti (12-coordinate) is 147 pm; radius of N atom is 75 pm. (3 points)arrow_forwardcan someone answer the questions and draw out the complete mechanismarrow_forward
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