Answer:
your answer is 12 hope it's correct answer
Identical electron configurations : K⁺ and Cl⁻
<h3>Further explanation </h3>
In an atom, there are levels of energy in the shell and sub-shell
This energy level is expressed in the form of electron configurations.
Charging electrons in the sub-shell uses the following sequence:
<em>1s², 2s², 2p⁶, 3s², 3p⁶, 4s², 3d¹⁰, 4p⁶, 5s², 4d¹⁰, 5p⁶, 6s², etc. </em>
S²⁻ : [Ne] 3s²3p⁶
Cl : [Ne] 3s²3p⁵
K⁺ : 1s² 2s² 2p⁶ 3s² 3p⁶
Cl⁻ : 1s² 2s² 2p⁶ 3s²3p⁶
S :[Ne] 3s²3p⁴
Ar : [Ne] 3s²3p⁶
Cl⁻ : 1s² 2s² 2p⁶ 3s²3p⁶
K : 1s² 2s² 2p⁶ 3s² 3p⁶4s¹
<h3>
Answer:</h3>
0.111 J/g°C
<h3>
Explanation:</h3>
We are given;
- Mass of the unknown metal sample as 58.932 g
- Initial temperature of the metal sample as 101°C
- Final temperature of metal is 23.68 °C
- Volume of pure water = 45.2 mL
But, density of pure water = 1 g/mL
- Therefore; mass of pure water is 45.2 g
- Initial temperature of water = 21°C
- Final temperature of water is 23.68 °C
- Specific heat capacity of water = 4.184 J/g°C
We are required to determine the specific heat of the metal;
<h3>Step 1: Calculate the amount of heat gained by pure water</h3>
Q = m × c × ΔT
For water, ΔT = 23.68 °C - 21° C
= 2.68 °C
Thus;
Q = 45.2 g × 4.184 J/g°C × 2.68°C
= 506.833 Joules
<h3>Step 2: Heat released by the unknown metal sample</h3>
We know that, Q = m × c × ΔT
For the unknown metal, ΔT = 101° C - 23.68 °C
= 77.32°C
Assuming the specific heat capacity of the unknown metal is c
Then;
Q = 58.932 g × c × 77.32°C
= 4556.62c Joules
<h3>Step 3: Calculate the specific heat capacity of the unknown metal sample</h3>
- We know that, the heat released by the unknown metal sample is equal to the heat gained by the water.
4556.62c Joules = 506.833 Joules
c = 506.833 ÷4556.62
= 0.111 J/g°C
Thus, the specific heat capacity of the unknown metal is 0.111 J/g°C
<span>The flame goes out on a burning match when sodium carbonate and hydrochloric
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