Answer:
None of these are correct, because there is no way to balance this equation, but I hope these steps help you figure out your answer.
Explanation:
Count out the single amounts of elements you have on both sides of the equation. To be balanced, you need to have the exact same for each element.
Before balanced Left side.
Cl-2
O-8
H-2
Before balanced right side.
H-1
Cl-1
O-3
That means we need to increase Hydrogen, Chlorine and Oxygen on the right for sure and see how that affects the equation. You can keep adding the Coefficients until the # of elements begin to match on each side.
(I tried to balance this equation, it doesn't work, there is too much on the reactants side for what the product is.)
Answer:
pH value of a solution depends on the concentration of hydrogen ions
(pH = -log[H+(aq)].
Hydrochloric acid is a strong acid, while ethanoic acid is a weak acid. Strong acids ionize completely in water (to give ions which includes H+(aq)), while weak acids only ionize partially in water.
Therefore, even if both hydrochloric acid and ethanoic acid are monobasic acids (each molecule can ionize completely to give 1 hydrogen ion), since hydrochloric acid ionizes completely in water and ethanoic acid does not ionize completely, the concentration of hydrogen ions in hydrochloric acid is higher than that of ethanoic acid, leading to a lower pH value for hydrochloric acid, while higher for ethanoic acid.
Answer: 77.4 mL
Explanation:
Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.
The combined gas equation is:
where,
= initial pressure of dry gas = (760 - 17.5) mmHg= 742.5 mm Hg
= final pressure of dry gas at STP = 760 mm Hg
= initial volume of dry gas = 85.0 mL
= final volume of dry gas at STP = ?
= initial temperature of dry gas =
= final temperature of dry gas at STP =
Now put all the given values in the above equation, we get the final volume of wet gas at STP
Volume of dry gas at STP is 77.4 mL.
Atomic mass of sodium is 22.9 g/mol
atomic mass of chloride is 35.45 g/mol
sodium chloride is 22.99 + 35.45= 58.44 g/mol