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almond37 [142]
3 years ago
7

Which elements can be magnetic? Question 35 options: A) Iron, Carbon, and Gold B) Cobalt, Nickel, and Iron C) Nickel, Gold, and

Aluminum
Chemistry
2 answers:
Komok [63]3 years ago
5 0

Answer:

A) Iron B) All of these metals are magnetic ones. Iron is the most magnetic one. C) Nickel

Explanation:

Magnetism varies on metals and non metals elements. There are five classes of Magnetism, and this classification is based on the magnetic moments of the atoms of element.

In crescent order:  

  1. Diamagnetic: Carbon, Gold Magnetic Moment below ranging from   small and negative susceptibility to magnetic.
  2. Paramagnetic Aluminum (small and positive  magnetic susceptibility) (1
  3. Antiferromagnetic  (small and positive magnetic susceptibility)
  4. Ferrimagnetic (mid and positive magnetic susceptibility)
  5. Ferromagnetic Iron, Nickel, Cobalt (mid and huge positive susceptibility).

Alexxandr [17]3 years ago
3 0

Answer:

B. is the correct answer. Plz mark as brainliest if helpful.

Explanation: Gold is not magnetic, so A. and B. are wrong

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This type of bonding can be described as a "cooperation" A. lonic B. metallic C. covalent​
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In a hypothetical atom, electron N transitions between energy levels, giving off orange light in the transition. In the same ato
Zielflug [23.3K]

Answer : Electron P has greater energy difference than the Electron N.

Explanation :  

Wavelength range of violet light = 400 - 500 nm

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The Planck's equation is,

E=\frac{h\times c}{\lambda}

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\lambda = wavelength of light

According to the Planck's equation, wavelength and energy follow inverse relation. As the wavelength increases, energy decreases.

From the given spectrum, the wavelength of violet light is less. We conclude that When electron P gives violet light on transition it means that energy difference between the energy level was high.

From the given spectrum, the wavelength of orange light is more. We conclude that When electron N  gives orange light on transition it means that energy difference between the energy level was low.

So, Electron P which gives violet light on transition has greater energy difference than the Electron N.


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3 years ago
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How would you separate a mixture of 4-tert-butylphenol and 2- chlorobenzoic acid? Explain the detailed steps including the choic
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The a mixture of 4-tert-butylphenol and 2- chlorobenzoic acid is separate by the by using  bicarbonate solvent.

When solution of bicarbonate is added in mixture of 4-tert-butylphenol and 2- chlorobenzoic acid then 2- chlorobenzoic acid form a carboxylate ion whereas 4-tert-butylphenol  is underacted and filtered out.

Since, only 2- chlorobenzoic acid which is acid is convert into its conjugate base by solution of bicarbonate in mixture of 4-tert-butylphenol and 2- chlorobenzoic acid .

However, phenol is less acidic than carboxylic acid. Both phenol and carboxylic acid is soluble in organic solvent . At that point as the phenol isolates as an oil, one needs to cool the blend in an ice shower to encourage crystallization

to learn more about 4-tert-butylphenol

brainly.com/question/5274368

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2 years ago
The enthalpy of solution (∆H) of KOH is -57.6 kJ/mol. If 3.66 g KOH is dissolved in enough water to make a 150.0 mL solution, wh
Wewaii [24]

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

The enthalpy of solution of KOH is -57.6 kJ/mol. We can calculate the heat released by the solution (Qr) of 3.66 g of KOH considering that the molar mass of KOH is 56.11 g/mol.

3.66g \times \frac{1mol}{56.11g} \times \frac{(-57.6kJ)}{mol} = -3.76 kJ

According to the law of conservation of energy, the sum of the heat released by the solution of KOH (Qr) and the heat absorbed by the solution (Qa) is zero.

Qr+Qa = 0\\\\Qa = -Qr = 3.76 kJ

150.0 mL of solution with a density of 1.02 g/mL were prepared. The mass (m) of the solution is:

150.0 mL \times \frac{1.02g}{mL}  = 153 g

Given the specific heat capacity of the solution (c) is 4.184 J/g・°C, we can calculate the change in the temperature (ΔT) of the solution using the following expression.

Qa = c \times m \times \Delta T\\\\\Delta T = \frac{Qa}{c \times m} = \frac{3.76 \times 10^{3}J  }{\frac{4.184J}{g.\° C }  \times 153g} = 5.87 \° C

When 3.66 g of KOH (∆Hsol = -57.6 kJ/mol) is dissolved in 150.0 mL of solution, it causes a temperature change of 5.87 °C.

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