All categories

3 years ago

How does temperature affect magnetism

1 answer:

3 0

Temperature affects magnetism by either strengthening or weakening a magnet's attractive force. A magnet subjected to heat experiences a reduction in its magnetic field as the particles within the magnet are moving at an increasingly faster and more sporadic rate.

You might be interested in

Please help quickly as possible thank you :)

WARRIOR [948]

Answer:

moving object transfers some to the stationary object causing it to move a it. remember that momentum is always conserved though - it is the same at the start before the event and after it.

Explanation:

4 0

3 years ago

What organic product would be formed from the reaction of 1-bromo-2-methylpropane (isobutyl bromide), (ch3)2chch2br, with mg, et

miskamm [114]

The reaction of 1-bromo-2-methylpropane (isobutyl bromide), $(CH_3)_2CHCH_2Br$ , with $Mg$ , and $Et_2O$ results in the formation of an adduct of formula $(CH_3)_2CHCH_2MgBr$ known as grignard reagent, an organometallic reagent, which on reacting with $H_3O^{+}$ results in the formation of isobutane.

The reaction is shown in the image.

4 0

3 years ago

2, classify the following molecules as polar or non polar.

Vitek1552 [10]

A. CH4= NON POLAR

B. CH3cl= POLAR

C. CO2= NON POLAR

D. H2O2= POLAR

E. BCl3= NON POLAR

F. H2S= SLIGHTLY POLAR

7 0

3 years ago

Different between a bird and a rat

Bess [88]

Explanation:

Rats and mice are both rodents, so look similar - the biggest difference is their size. Rats are larger and heavier while mice have smaller slender bodies. Mice also have long slender tails (for their body size) covered in hair compared to rat tails which are shorter, thicker and hairless

5 0

2 years ago

Calculate the percent composition of the following:<br><br> A. VO3<br><br> B. V2O5

tino4ka555 [31]

A) Answer is: 51.48% V and 48.52% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(VO₃) = 50.94 + 3 · 16.

Ar(VO₃) = 98.94; molecular weight of vanadium (VI) oxide.

ω(V) = Ar(V) ÷ Ar(VO₃) · 100%.

ω(V) = 51.48%; the percent composition of vanadium.

ω(O) = 100% - 51.48%.

ω(O) = 48.52%; the percent composition of oxygen.

B) Answer is: 67.80% V and 32.20% O.

Ar(V) = 50.94; atomic weight of vanadium.

Ar(O) = 16; atomic weight of oxygen.

Ar(V₂O₅) = 2 · 50.94 + 5 · 16.

Ar(V₂O₅) = 149.88; molecular weight of vanadium (V) oxide.

ω(V) = 2 · Ar(V) ÷ Ar(V₂O₅) · 100%.

ω(V) = 67.80%; the percent composition of vanadium.

ω(O) = 100% - 67.80%.

ω(O) = 32.20%; the percent composition of oxygen.

4 0

3 years ago