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1 answer:
⭐ Elements in which the last electron enters any one of the five d-oribitals of their respective penultimate shells are called as <u>d-block</u><u> </u><u>elem</u><u>ents</u> .
⭐ But the last electron of Zn , Cd , Hg and Cn enters in the s-oribital of their respective ultimate shells rather than the d-oribitals of their respective penultimate shells . Therefore, these elements cannot be regarded as d-block elements .
☃️ But properties of these elements resemble to the d-block elements rather than s-block elements .
☃️ Therefore, to make the study of periodic classification of elements more rational, they are studied along with d-block elements .
✍️ Thus <u>on the basis of properties</u> all transition elements are d- block elements, but <u>on the basis of electronic configuration</u> all d -block elements are not transition elements .
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Answer:
The temperature of the air at this given elevation will be 53.32425°C
Explanation:
We can calculate the final temperature through the combined gas law. Therefore we will need to know 1 ) The initial volume, 2 ) The initial temperature, 3 ) Initial Pressure, 4 ) Final Volume, 5 ) Final Pressure.
Initial Volume = 1.2 L ; Initial Temperature = 25°C = 298.15 K ; Initial pressure = 1.0 atm ; Final Volume = 1.8 L ; Final pressure = 0.73 atm
We have all the information we need. Now let us substitute into the following formula, and solve for the final temperature ( T ),
PV
/ T
= P
V
/ T
,
T = P
V
T
/ P
V
,
T = 0.73 atm
1.8 L
298.15 K / 1 atm
1.2 L = ( 0.73
1.8
298.15 / 1
1.2 ) K = 326.47425 K,
T = 326.47425 K = 53.32425 C
Given that 4.50 dm³ of Pb(NO₃)₂ is cooled from 70 °C to 18 °C, the
amount amount of solute that will be deposited is 1,927.413 grams.
<h3>How can the amount of solute deposited be found?</h3>
The volume of water 1.33 dm³ of water 70 °C.
The number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water at 70 °C = 2.25 moles
At 18 °C, the number of moles of Pb(NO₃)₂ that saturates 1.33 dm³ of water = 0.53 moles
Therefore;
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C is therefore;
1.33 dm³ contains 2.25 moles.
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 70 °C ≈ 7.613 moles
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C is therefore;
1.33 dm³ contains 0.53 moles
Number of moles of Pb(NO₃)₂ in 4.50 dm³ at 18 °C ≈ 1.79 moles
The number of moles that precipitate out = The amount of solute deposited
Which gives;
Amount of solute deposited = 7.613 moles - 1.79 moles = 5.823 moles
The molar mass of Pb(NO₃)₂ = 207 g + 2 × (14 g + 3 × 16 g) = 331 g
The molar mass of Pb(NO₃)₂ = 331 g/mol
The amount of solute deposited = Number of moles × Molar mass
Which gives;
The amount of solute deposited = 5.823 moles × 331 g/mol =<u> 1,927.413 g </u>
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