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Bingel [31]
2 years ago
5

What are the physical and chemical properties of sodium?

Physics
1 answer:
Hitman42 [59]2 years ago
8 0

Answer:

Physical Properties of Sodium

Atomic number 11

Melting point 97.82°C (208.1°F)

Boiling point 881.4°C (1618°F)

Volume increase on melting 2.70%

Latent heat of fusion 27.0 cal/g

Lenntech Water treatment & purification

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Home Periodic table Elements Sodium

Sodium - Na

Chemical properties of sodium - Health effects of sodium - Environmental effects of sodium

Atomic number

11

Atomic mass

22.98977 g.mol -1

Electronegativity according to Pauling

0.9

Density

0.97 g.cm -3 at 20 °C

Melting point

97.5 °C

Boiling point

883 °C

Vanderwaals radius

0.196 nm

Ionic radius

0.095 (+1) nm

Isotopes

3

Electronic shell

[Ne] 3s1

Energy of first ionisation

495.7 kJ.mol -1

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A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
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Answer:

The average emf that will be induced in the wire loop during the extraction process is 37.9 V

Explanation:

The average emf induced can be calculated from the formula

Emf = -N\frac{\Delta \phi}{\Delta t}

Where N is the number of turns

\Delta \phi is the change in magnetic  flux

\Delta t is the time interval

The change in magnetic flux is given by

\Delta \phi = \phi _{f} - \phi _{i}

Where \phi _{f} is the final magnetic flux

and \phi _{i} is the initial magnetic flux

Magnetic flux is given by the formula

\phi = BAcos(\theta)

Where B is the magnetic field

A is the area

and \theta is the angle between the magnetic field and the area.

Initially, the magnetic field and the area are pointed in the same direction, that is, \theta = 0^{o}

From the question,

B = 1.5 T

and radius = 15.0 cm = 0.15 m

Since it is a circular loop of wire, the area is given by

A = \pi r^{2}

∴ A = \pi (0.15)^{2}

A = 0.0225\pi

∴\phi_{i}  = (1.5)(0.0225\pi)cos(0^{o} )

\phi_{i}  = (1.5)(0.0225\pi)

( NOTE: cos (0^{o}) = 1 )

\phi_{i}  = 0.03375\pi Wb

For \phi_{f}

The field pointed upwards, that is \theta = 90^{o}. Since cos (90^{o}) = 0

Then

\phi_{f} = 0

Hence,

\Delta \phi = 0- 0.03375\pi

\Delta \phi = - 0.03375\pi

From the question

\Delta t = 2.8 ms = 2.8 \times 10^{-3} s

Here, N = 1

Hence,

Emf = -N\frac{\Delta \phi}{\Delta t} becomes

Emf = -(1)\frac{-0.03375\pi}{2.8 \times 10^{-3} }

Emf = 37.9 V

Hence, the average emf that will be induced in the wire loop during the extraction process is 37.9 V.

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