Base + acid = salt + water
Example :
<span>HCl + NaOH = NaCl + H</span>₂<span>O
</span>
Answer A
hope this helps!
Answer:
you answers are sodium and chloride oxygen
Explanation:
sodium
chloride oxygen
Answer:
72.8 % (But verify explanation).
Explanation:
Hello,
In this case, with the following obtained results, the percent error is computed as follows:
Volume of vinegar= 7.0 mL
Volume of NaOH= (7+6.6+6.4)/3= 6.7 mL
Used concentration of NaOH= 1.5M
Concentration of acetic acid= (concentration of NaOH*volume of NaOH)/volume of vinegar= (6.7mL*1.5M)/7.0M= 1.44M
Assuming we have 100 mL (0.100L) of vinegar, moles of acetic acid in vinegar = 1.44M x 0.100 L= 0.144 mol
Mass of acetic acid in 100g of vinegar = 0.144 mol x 60.0g/mol= 8.64 g
% of acetic acid in vinegar=8.64 %
% error in percentage of acetic acid = [(8.64% - 5.0%)/5.0] x 100=72.8 %
Clearly, this result depend on your own measurements, anyway, you can change any value wherever you need it.
Regards.
Answer:
pH = 0.35
Explanation:
For a strong acid, all of the acid dissociates into H3O+, and pH = -log[H3O+], where [H3O+] = [HClO3] = 0.45 M.
Question:
Zinc metal is added to hydrochloric acid to generate hydrogen gas and is collected over a liquid whose vapor pressure is the same as pure water at 20.0 degrees C (18 torr). The volume of the mixture is 1.7 L and its total pressure is 0.987 atm. Determine the number of moles of hydrogen gas present in the sample.
A. 0.272 mol
B. 0.04 mol
C. 0.997 mol
D. 0.139 mol
E. 0.0681 mol
Answer:
The correct option is;
E. 0.0681 mol
Explanation:
The equation for the reaction is
Zn + HCl = H₂ + ZnCl₂
Vapor pressure of the liquid = 18 torr = 2399.803 Pa
Total pressure of gas mixture H₂ + liquid vapor = 0.987 atm
= 100007.775 Pa
Therefore, by Avogadro's law, pressure of the hydrogen gas is given by the following equation
Pressure of H₂ = 100007.775 Pa - 2399.803 Pa = 97607.972 Pa
Volume of H₂ = 1.7 L = 0.0017 m³
Temperature = 20 °C = 293.15 K
Therefore,

Therefore, the number of moles of hydrogen gas present in the sample is n ≈ 0.0681 moles.