Answer:
Aluminium.
Explanation:
The above electronic configuration can be written in a simplified form as shown below:
1s² 2s²2p⁶ 3s²3p¹
Next, we shall determine the number of electrons in the atom of the element as follow:
Number electron = 2 + 2 + 6 + 2 + 1
Number of electron = 13
Next, we shall determine the number of protons.
Since the element is in its neutral state,
The number of electrons and protons are equal i.e
Proton = Electron
Number of electron = 13
Proton = Electron = 13
Proton = 13
Next, we shall determine the atomic number of the element.
The atomic number of an element is simply the number of protons in the atom of the element i.e
Atomic number = proton number
Proton = 13
Atomic number = 13
Comparing the atomic number of the element with those in the periodic table, the element with the above electronic configuration is aluminium since no two elements have the same atomic number.
The amount of heat energy needed to convert 400 g of ice at -38 °C to steam at 160 °C is 1.28×10⁶ J (Option D)
<h3>How to determine the heat required change the temperature from –38 °C to 0 °C </h3>
- Mass (M) = 400 g = 400 / 1000 = 0.4 Kg
- Initial temperature (T₁) = –25 °C
- Final temperature (T₂) = 0 °
- Change in temperature (ΔT) = 0 – (–38) = 38 °C
- Specific heat capacity (C) = 2050 J/(kg·°C)
- Heat (Q₁) =?
Q = MCΔT
Q₁ = 0.4 × 2050 × 38
Q₁ = 31160 J
<h3>How to determine the heat required to melt the ice at 0 °C</h3>
- Mass (m) = 0.4 Kg
- Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
- Heat (Q₂) =?
Q = mL
Q₂ = 0.4 × 334000
Q₂ = 133600 J
<h3>How to determine the heat required to change the temperature from 0 °C to 100 °C </h3>
- Mass (M) = 0.4 Kg
- Initial temperature (T₁) = 0 °C
- Final temperature (T₂) = 100 °C
- Change in temperature (ΔT) = 100 – 0 = 100 °C
- Specific heat capacity (C) = 4180 J/(kg·°C)
- Heat (Q₃) =?
Q = MCΔT
Q₃ = 0.4 × 4180 × 100
Q₃ = 167200 J
<h3>How to determine the heat required to vaporize the water at 100 °C</h3>
- Mass (m) = 0.4 Kg
- Latent heat of vaporisation (Hv) = 2260 KJ/Kg = 2260 × 1000 = 2260000 J/Kg
- Heat (Q₄) =?
Q = mHv
Q₄ = 0.4 × 2260000
Q₄ = 904000 J
<h3>How to determine the heat required to change the temperature from 100 °C to 160 °C </h3>
- Mass (M) = 0.4 Kg
- Initial temperature (T₁) = 100 °C
- Final temperature (T₂) = 160 °C
- Change in temperature (ΔT) = 160 – 100 = 60 °C
- Specific heat capacity (C) = 1996 J/(kg·°C)
- Heat (Q₅) =?
Q = MCΔT
Q₅ = 0.4 × 1996 × 60
Q₅ = 47904 J
<h3>How to determine the heat required to change the temperature from –38 °C to 160 °C</h3>
- Heat for –38 °C to 0°C (Q₁) = 31160 J
- Heat for melting (Q₂) = 133600 J
- Heat for 0 °C to 100 °C (Q₃) = 167200 J
- Heat for vaporization (Q₄) = 904000 J
- Heat for 100 °C to 160 °C (Q₅) = 47904 J
- Heat for –38 °C to 160 °C (Qₜ) =?
Qₜ = Q₁ + Q₂ + Q₃ + Q₄ + Q₅
Qₜ = 31160 + 133600 + 167200 + 904000 + 47904
Qₜ = 1.28×10⁶ J
Learn more about heat transfer:
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Solving part-1 only
#1
KMnO_4
- Transition metal is Manganese (Mn)
#2
Actually it's the oxidation number of Mn
Let's find how?
- x is the oxidation number
#3
- Purple as per the color of potassium permanganate
#4
Answer : Electron P has greater energy difference than the Electron N.
Explanation :
Wavelength range of violet light = 400 - 500 nm
Wavelength range of orange light = 600 - 700 nm
The Planck's equation is,
where,
E = energy of light
c = speed of light
= 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.
Answer:A
Explanation:
A nuclear reaction is balanced by ensuring that the Masses and charges of te reactants and products are exactly balanced on the left and right hand side of the reaction equation. If there are 60 mass units on the LHS and manganese has only 56 mass units then four mass units are left. If there is no charge on a neutron and there is a charge of 27 on the cobalt, then two charges are left. Four mass units and a charge of +2 corresponds to a helium which is actually an alpha particle.
