I believe it is the Great Plains in Nebraska.
Answer:
q1 = mCpΔT
= 18.016g × 1.84J/g.K × (418.15-373.15)
= 1491.72 J
q2 = n×ΔH vap
= 1mol ×44.0kJ/mol
= 44KJ
∴ qtotal = q1+ q2
= 1.498kJ + 44.0kJ
= 45.498KJ
Explanation: The heat flow can be separated into steps.all that is being observed at a constant pressure,the heat flow is equal to the enthalpy.
Answer:
10 L of CO₂.
Explanation:
The balanced equation for the reaction is given below:
2CO + O₂ —> 2CO₂
From the balanced equation above,
2 L of CO reacted to produce 2 L of CO₂.
Finally, we shall determine the volume of CO₂ produced by the reaction of 10 L CO. This can be obtained as follow:
From the balanced equation above,
2 L of CO reacted to produce 2 L of CO₂.
Therefore, 10 L of CO will also react to produce 10 L of CO₂.
Thus, 10 L of CO₂ were obtained from the reaction.
POH = 14 - pH
pOH = -log [OH-]
Answer : The ratio of the protonated to the deprotonated form of the acid is, 100
Explanation : Given,

pH = 6.0
To calculate the ratio of the protonated to the deprotonated form of the acid we are using Henderson Hesselbach equation :
![pH=pK_a+\log \frac{[Salt]}{[Acid]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BSalt%5D%7D%7B%5BAcid%5D%7D)
![pH=pK_a+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=pH%3DpK_a%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
Now put all the given values in this expression, we get:
![6.0=8.0+\log \frac{[Deprotonated]}{[Protonated]}](https://tex.z-dn.net/?f=6.0%3D8.0%2B%5Clog%20%5Cfrac%7B%5BDeprotonated%5D%7D%7B%5BProtonated%5D%7D)
As per question, the ratio of the protonated to the deprotonated form of the acid will be:
Therefore, the ratio of the protonated to the deprotonated form of the acid is, 100