Answer: acceleration
a = 5.36 m/s²
Explanation: solution attached:
Convert first 60 mi/h to m/s
Use the formula for acceleration
a = vf - vi /t
I believe the correct answer from the choices listed above is the third option. A ball falling through the air has <span>both potential energy because it still has a distance to move before it hits the ground and kinetic energy because it is moving. Hope this answers the question. Have a nice day.</span>
Answer: 100kPa
Explanation:
P1 = 3.00 x 10² kPa , P2 =?
T1 = 30°C = 30 +273 = 303k
T2 = —172°C = —172 + 273 = 101k
P1/T1 = P2/T2
3 x 10² / 303 = P2 / 101
P2 = (3 x 10² / 303) x 101
P2 = 100kPa
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
The silver ion concentration in saturated solution of silver (i) phosphate is calculated as follows.
write the equation for dissociation of silver (i) phosphate
that is Ag3PO4 (s) = 3Ag^+(aq) + PO4 ^3-(aq)
let the concentration of the ion be represented by x
ksp is therefore= (3x^3 )(x) = 1.8 x10 ^-18
27 x^3 (x) = 1.8 x10^-18
27x^4 = 1.8 x10^-18 divide both side 27
X^4 = 6.67 x10 ^-20
find the fourth root x = 1.6 x10 ^-5m
the concentration of silver ion is therefore = 3 x (1.6 x10^-5) = 4.8 x10^-5m
