How are electromagnets temporary?

2 answers:

8 0

Such magnets are called temporary magnets. An electromagnet is a temporary magnet. It is made by winding a coil of insulated wire round a soft iron core. When a current is passed through the coil, the magnetic field produced by the current magnetises the soft iron core. The soft iron core loses all the magnetism when the current is switched off.

So A

Tju [1.3M]3 years ago

5 0

The answer (to this question) is A

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The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

<u>Answer:</u> The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

<u>For $_{77}^{191}\textrm{Ir}$ isotope:</u>

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

<u>For $_{77}^{193}\textrm{Ir}$ isotope:</u>

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

8 0

3 years ago

A breathing mixture used by deep-sea divers contains helium, oxygen, and carbon dioxide. what is the partial pressure of oxygen

pentagon [3]

Pressure of the Mixture = 101.4 kpa
Pressure of the helium phe = 82.5 kpa
Pressure of CO2 pco2 = 0.4 kpa.
Partial pressure of oxygen = Pressure of the Mixutre - (Pressure of the
helium + Pressure of CO2)
Partial pressure of oxygen = 101.4 - 82.5 - 0.4
Partial pressure of oxygen = 18.5 kPa

7 0

3 years ago

Read 2 more answers

Chemistry: Gas Laws Problems!

Ronch [10]

Answer : The final volume at STP is, 1000 L

Explanation :

According to the Boyle's, law, the pressure of the gas is inversely proportional to the volume of gas at constant temperature and moles of gas.

$P\propto \frac{1}{V}$