PV=nRT
P=nRT/V
P=[(0.650mol)(0.08206)(298K)]/(0.750L)=21.2atm
Answer:
The products will be;
CO₂ + H₂O + NaC₂H₃O₂
Explanation:
We are given;
Two reactants NaHCO₃ and HC₂H₃O₂
We are supposed to determine the products;
- We need to know that hydrogen carbonates reacts with acids to give water, carbon dioxide and a salt as the products.
Therefore;
- In this case, sodium hydrogen carbonates (baking soda) reacts with acetic acid to form water, carbon dioxide and sodium acetate.
- The equation for the reaction is;
NaHCO₃ + HC₂H₃O₂ → CO₂ + H₂O + NaC₂H₃O₂
- Therefore, the products of the complete reaction between sodium hydrogen carbonate and acetic acid are CO₂ + H₂O + NaC₂H₃O₂
To calculate the new pressure, we can use Boyle’s law to relate these two scenarios (Boyle’s law is used because the temperature is assumed to remain constant). Boyle’s law is:
P1V1 = P2V2,
Where “P” is pressure and “V” is volume. The pressure and volume of the first scenario is 215 torr and 51 mL, respectively, and the second scenario has a volume of 18.5 L (18,500 mL) and the unknown pressure - let’s call that “x”. Plugging these into the equation:
(215 torr)(51 mL) =(“x” torr)(18,500 mL)
x = 0.593 torr
The final pressure exerted by the gas would be 0.593 torr.
Hope this helps!
Answer:
0.056moles HF and 0.70M
Explanation:
When a strong acid is added to a buffer, the acid reacts with the conjugate base.
In the system, NaF and HF, weak acid is HF and conjugate base is NaF. The reaction of NaF with HCl (Strong acid) is:
NaF + HCl → HF + NaCl
Initial moles of NaF and HF in 60.0mL of solution are:
NaF:
0.0600L × (0.80mol / L)= 0.048 moles NaF
HF:
0.0600L × (0.80mol / L)= 0.048 moles HF
Then, the added moles of HCl are:
0.0200L × (0.40mol / L) = 0.008 moles HCl.
Thus, after the reaction, moles of HF produced are 0.008 moles + the initial 0.048moles of HF, moles of HF are:
<em>0.056moles HF</em>
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In 20.0mL + 60.0mL = 80.0mL = 0.0800L, molarity of HF is:
0.056mol HF / 0.0800L = <em>0.70M</em>
