Answer:
HF
H₂S
H₂CO₃
NH₄⁺
Explanation:
<em>Which acid in each of the following pairs has the stronger conjugate base?</em>
According to Bronsted-Lowry acid-base theory, <em>the weaker an acid, the stronger its conjugate acid</em>. Especially for weak acids, pKa gives information about the strength of such acid. <em>The higher the pKa, the weaker the acid.</em>
<em />
- Of the acids HCl or HF, the one with the stronger conjugate base is HF because it is a weak acid.
- Of the acids H₂S or HNO₂, the one with the stronger conjugate base is H₂S because it is a weaker acid. pKa (H₂S) = 7.04 > pKa (HNO₂) = 3.39
- Of the acids H₂CO₃ or HClO₄, the one with the stronger conjugate base is H₂CO₃ because it is a weak acid.
- Of the acids HF or NH₄⁺, the one with the stronger conjugate base is NH₄⁺ because it is a weaker acid. pKa (HF) = 3.17 < pKa (NH₄⁺) = 9.25
<span>294400 cal
The heating of the water will have 3 phases
1. Melting of the ice, the temperature will remain constant at 0 degrees C
2. Heating of water to boiling, the temperature will rise
3. Boiling of water, temperature will remain constant at 100 degrees C
So, let's see how many cal are needed for each phase.
We start with 320 g of ice and 100 g of liquid, both at 0 degrees C. We can ignore the liquid and focus on the ice only. To convert from the solid to the liquid, we need to add the heat of fusion for each gram. So multiply the amount of ice we have by the heat of fusion.
80 cal/g * 320 g = 25600 cal
Now we have 320 g of ice that's been melted into water and the 100 g of water we started with, resulting in 320 + 100 = 420 g of water at 0 degrees C. We need to heat that water to 100 degrees C
420 * 100 = 42000 cal
Finally, we have 420 g of water at the boiling point. We now need to pump in an additional 540 cal/g to boil it all away.
420 g * 540 cal/g = 226800 cal
So the total number of cal used is
25600 cal + 42000 cal + 226800 cal = 294400 cal</span>
C. NaOH ammmonia is also an base but not as strong as NaOH
Answer:
hello your question is incomplete the options are missing
Determine the resultant velocity for the plane when it is travelling
i) To the east
ii) To the west
answer :i) 270 i
ii) -150 i
Explanation:
velocity of Airplane = 210 m/s
wind velocity = 60 m/s to the east
The resultant velocity for the plane when it is travelling
let the velocity of the wind = V2
velocity of the plane = v1
i) The resultant velocity for the plane when travelling to the east
Vr = V2 i + V1 i
Vr= 60i + 210i = 270i
ii) resultant velocity when the plane is travelling to the west
Vr = - V1 i + V2i
= -210i + 60 i = -150 i
..........The answer is B
