Answer :
The pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.
The pressure of carbon dioxide in the atmosphere in atm is 
.
Explanation :
The conversion used for pressure from torr to mmHg is:
1 torr = 1 mmHg
The conversion used pressure from torr to atm is:
1 atm = 760 torr
or,
As we are given the pressure of carbon dioxide in the atmosphere 0.239 torr. Now we have to determine the pressure of carbon dioxide in the atmosphere in mmHg and atm.
<u>Pressure in mmHg :</u>
As, 
So, 
Thus, the pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.
<u>Pressure in atm:</u>
As,
So, 
Thus, the pressure of carbon dioxide in the atmosphere in atm is .
Answer:
a. pH = 13.50
b. pH = 13.15
Explanation:
Hello!
In this case, since the undergoing chemical reaction between KOH and HBr is:
As they are both strong. In such a way, since the initial analyte is the 25.00 mL solution of 0.320-M KOH, we first compute the pOH it has, considering that all the KOH is ionized in potassium and hydroxide ions:
![pOH=-log([OH^-])=-log(0.320)=0.50](https://tex.z-dn.net/?f=pOH%3D-log%28%5BOH%5E-%5D%29%3D-log%280.320%29%3D0.50)
Thus, the pH is:
Which is the same answer for a and b as they ask the same.
Moreover, once 5.00 mL of the HBr is added, we need to compute the reacting moles of each substance:
It means that since there are more moles of KOH, we need to compute the remaining moles after those 0.00375 moles of acid consume 0.00375 moles of base because they are in a 1:1 mole ratio:
Next, we compute the resulting concentration of hydroxide ions (equal to the concentration of KOH) in the final solution of 30.00 mL (25.00 mL + 5.00 mL):
![[OH^-]=[KOH]=\frac{0.00425mol}{0.03000L}=0.142M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%5BKOH%5D%3D%5Cfrac%7B0.00425mol%7D%7B0.03000L%7D%3D0.142M)
So the pOH and the pH turn out:
Best regards!
Answer:
118.776 mmHg
Explanation:
The equation of the reaction is;
C4H10(g) + 13/2 O2(g) ------> 4CO2(g) + 5H20(g)
Now the mole ratio according to the balanced reaction equation is;
1 : 6.5 : 4 : 5
Hence, the total number of moles present = 1 + 6.5 + 4 + 5 = 16.5 moles
Mole fraction of water vapour = 5/16.5 = 0.303
We also know that;
Partial pressure= mole fraction * total pressure
Partial pressure of H20(g) = 0.303 * 392 mmHg = 118.776 mmHg
Answer:
There is 17.1 kJ energy required
Explanation:
Step 1: Data given
Mass of ethanol = 322.0 grams
Initial temperature = -2.2 °C = 273.15 -2.2 = 270.95K
Final temperature = 19.6 °C = 273.15 + 19.6 = 292.75 K
Specific heat capacity = 2.44 J/g*K
Step 2: Calculate energy
Q = m*c*ΔT
⇒ m = the mass of ethanol= 322 grams
⇒ c = the specific heat capacity of ethanol = 2.44 J/g*K
⇒ ΔT = T2 - T1 = 292.75 - 270.95 = 21.8 K
Q = 322 * 2.44 * 21.8 = 17127.8 J = 17.1 kJ
There is 17.1 kJ energy required
Answer:
This is a single replacement reaction because I replaces Br.
