Answer:
B and C
Explanation:
I think the answer correct is C because you never know in what temperature the block of ice is going to melt but if it says select all that apply its possible that B might be useful.
Answer:
6,000kg/m3
Explanation:
6.00g/1cm3 x 1kg/1000g x 1cm3/0.000001m3
= 6.00kg/0.001m3
= 6,000kg/m3
Answer:
VP (solution) = 171.56 mmHg
Explanation:
Vapor pressure of pure solvent(P°) - Vapor pressure of solution (P') = P° . Xm
Let's replace the data:
173.11 mmHg - P' = 173.11 mmHg . Xm
Let's determine the Xm (mole fraction for solute)
Mole fraction for solute = Moles of solute / Total moles
Total moles = Moles of solute + moles of solvent.
Let's determine the moles
Moles of solvent → 623.4 g / 119.4 g/mol = 5.22 moles
Moles of solute → 9.322 g / 180.1 g/mol = 0.052 moles
Total moles = 0.052 + 5.22 = 5.272 moles
Xm = 0.052 moles / 5.272 moles = 0.009 → 9/1000
173.11 mmHg - P' = 173.11 mmHg . 9/1000
P' = - (173.11 mmHg . 9/1000 - 173.11 mmHg)
P' = 171.56 mmHg
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
