The term for a push or pull is called Force.
Answer:
One: <u>Selenium</u> is Paramagnetic
Explanation:
Those compounds which have unpaired electrons are attracted towards magnet. This property is called as paramagnetism. Lets see why remaining are not paramagnetic.
Electronic configuration of Scandium;
Sc = 21 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹
Sc³⁺ = 1s², 2s², 2p⁶, 3s², 3p⁶
Hence in Sc³⁺ there is no unpaired electron.
Electronic configuration of Bromine;
Br = 35 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁵
Br⁻ = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶
Hence in Br⁻ there is no unpaired electron.
Electronic configuration of Magnesium;
Mg = 12 = 1s², 2s², 2p⁶, 3s²
Mg²⁺ = 1s², 2s², 2p⁶
Hence in Mg²⁺ there is no unpaired electron.
Electronic configuration of selenium;
Se = 34 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁴
Or,
Se = 34 = 1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4px², 4py¹, 4pz¹
Hence in Se there are two unpaired electrons hence it is paramagnetic in nature.
<span>Pitch is sometimes defined as the fundamental frequency of a sound wave (i.e. generally, the lowest frequency in a given sound wave). For most practical purposes, this is fine, and pitch and frequency can be thought of as equivalent. On the other hand, for most practical purposes, amplitude can be thought of as volume.However, technically, pitch (and volume) are human perceptions. Thus, our perception of pitch and volume are not solely based on frequency and amplitude respectively, but are based on a combination of both (and even other factors). Frequency overwhelming dictates perceived pitch, but amplitude also does have some small, small effect on our pitch perception, especially when it is very large. For example, a very loud sound can have a different <span>perceived </span>pitch than you would predict from its frequency alone.That all being said, usually these effects are negligible, and pitch can be thought of as equivalent to fundamental frequency.
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Answer:
Vapor pressure of solution → 151.1 Torr
Option 2.
Explanation:
Raoult's Law is relationed to colligative property about vapor pressure. A determined solute, can make, the vapor pressure of solution decreases.
ΔP = P° . Xm
where Xm is the mole fraction of solute, P° (vapor pressure of pure solvent)
and ΔP = Vapor pressure of pure solvent - Vapor pressure of solution.
In order to determine the vapor pressure of solution, we need to determine, the vapor pressure of B and A in the solution
B's pressure = P° B . Xm
When we add A to B, A works as the solute and B, as the solvent.
Vapor pressure of pure B is 135 torr. (P° B)
In order to determine, the Xm, we use the moles of A and B
Xm = 5.3 mol of B / (1.28 + 5.3) → 0.806
B's pressure = 135 Torr . 0.806 → 108.81 Torr
If mole fraction of B is 0.806, mole fraction for A (solute) will be (1 - 0.806)
A's pressure = 218 Torr . 0.194 → 42.3 Torr
Vapor pressure of solution is sum of vapor pressures of solute + solvent.
Vapor pressure of solution = 42.3 Torr + 108.81 Torr → 151.1 Torr