The answer is the 3rd one down I think
Answer:
When we look at an arbitrary point in the sky, away from the sun, we see only the light that was redirected by the atmosphere into our line of sight. Because that occurs much more often for blue light than for red, the sky appears blue. Violet light is actually scattered even a bit more strongly than blue.
Explanation:
Thank you for posting your question here. Below is the solution:
HNO3 --> H+ + NO3-
<span>HNO3 = strong acid so 100% dissociation </span>
<span>** one doesn't need to find the molarity of water since it is the solvent </span>
<span>0M HNO3 </span>
<span>1x10^-6M H3O+ </span>
<span>1x10^-6M NO3- </span>
<span>1x10^-8M OH-.....the Kw = 1x10^-14 = [H+][OH-] </span>
<span>you have 1x10^-6M H+ so, 1x10^-14 / 1x10^-6 = 1x10^-8M OH- </span>
<span>1x10^-6 Ba(OH)2 = strong base, 100% dissociation </span>
<span>1x10^-6M Ba2+ </span>
<span>2x10^-6M OH- since there are 2 OH- / 1 Ba2+ </span>
<span>0M Ba(OH)2 </span>
<span>5x10^-9M H3O+</span>
Answer:
Option D. 4.02 kJ
Explanation:
A simple calorimetry problem
Q = m . C . ΔT
ΔT = Final T° - Initial T°
C = Specific heat capacity
m = mass
Let's replace the data
Q = 125 g . 2.42 J/g∘C . (34.8°C -21.5 °C)
Q= 4023.25 J
We must convert the answer to kJ
4023.25 J . 1kJ /1000 =4.02kJ
Answer:
Ka = 4.76108
Explanation:
- CO(g) + 2H2(g) ↔ CH3OH(g)
∴ Keq = [CH3OH(g)] / [H2(g)]²[CO(g)]
[ ]initial change [ ]eq
CO(g) 0.27 M 0.27 - x 0.27 - x
H2(g) 0.49 M 0.49 - x 0.49 - x
CH3OH(g) 0 0 + x x = 0.11 M
replacing in Ka:
⇒ Ka = ( x ) / (0.49 - x)²(0.27 - x)
⇒ Ka = (0.11) / (0.49 - 0.11)² (0.27 - 0.11)
⇒ Ka = (0.11) / (0.38)²(0.16)
⇒ Ka = 4.76108
