Answer:
![[Ar] 3d^{2} 4s^{2}](https://tex.z-dn.net/?f=%5BAr%5D%203d%5E%7B2%7D%204s%5E%7B2%7D)
Explanation:
You can look at the periodic table and figure out the electron config.
Answer:
Student 3
Explanation:
This question lets us know something about how the density of a gas varies with temperature.
When a gas is heated, its molecules spread out and expand. When this happens, the volume of the gas increases. Remember that density is defined as mass/volume. Thus as the volume increases, the density of the gas decreases.
Therefore, the carbon dioxide rose up because the heat expanded the gas and it became less dense.
Part A
75.0 mL of 0.10 M HF; 55.0 mL of 0.15 M NaF
This combination will form a buffer.
Explanation
Here, weak acid HF and its conjugate base F- is available in the solution
Part B
150.0 mL of 0.10 M HF; 135.0 mL of 0.175 M HCl
This combination cannot form a buffer.
Explanation
Here, moles of HF = 0.15 x 0.1 = 0.015 moles
Moles of HCl = 0.135 x 0.175 = 0.023
Since HCl is a strong acid and the number of HCl is higher than HF. This prevents the dissociation of HF and the conjugate base F- will not be available in the solution
Part C
165.0 mL of 0.10 M HF; 135.0 mL of 0.050 M KOH
This combination will form a buffer.
Explanation
Moles of HF = 0.165 x 0.1 = 0.0165 moles
Moles of KOH = 0.135 x 0.05 = 0.00675 moles
Moles of KOH is not sufficient for the complete neutralization of HF. Thus weak acid HF and its conjugate base F- is available in the solution and form a buffer
Part D
125.0 mL of 0.15 M CH3NH2; 120.0 mL of 0.25 M CH3NH3Cl
This combination will form a buffer
Explanation
Here, weak acid CH3NH3+ and its conjugate base CH3NH2 is available in the solution and form a buffer
Part E
105.0 mL of 0.15 M CH3NH2; 95.0 mL of 0.10 M HCl
This combination will form a buffer
Explanation
Moles of CH3NH2 = 0.105 x 0.15 = 0.01575 moles
Moles of HCl = 0.095 x 0.1 = 0.0095 moles
Thus the HCl completely reacts with CH3NH2 and converts a part of the CH3NH2 to CH3NH3+. This results weak acid CH3NH3+ and its conjugate base CH3NH2 is in the solution and form a buffer
These are properties of alkaline base.
<span>D.NaOH</span>
The new volume at standard pressure of 1 atm is 21294 liters.
Explanation:
Data given:
Initial volume of the gas V1 = 338 liters
initial pressure on the gas P1 = 63 atm
standard pressure as P2 = 1 atm
Final volume at standard pressure V2 =?
The data given shows that Boyle's law equation is to used:
P1V1 = P2V2
rearranging the equation to calculate V2,
V2 = 
Putting the values in the equation:
V2 = 
= 21294 L
as the pressure on the gas is reduced to 1 atm the volume of the gas increased incredibly to 21294 litres.
