The enthalpy of an intermediate step should be manipulated when used to produce an overall equation by using the Hess's law. You could multiply the enthalpy by -1 if this equation is reversed in theory.
Answer:
b. lithium
Explanation:
furthest to the left in second period
Answer 8.0 L.
2.0L / 5.0 moles = x / 20.0 => x = 20 / 5 * 2 = 8
The answer to your question is false
Answer:
Explanation:
A buffer is defined as an aqueous mixture of a weak acid and its conjugate base or vice versa.
In the systems:
H₂CO₃(aq) and KHCO₃(aq): Carbonic acid, H₂CO₃, is a weak acid that, in solution with its conjugate pair, HCO₃⁻ make a <em>buffer system.</em>
NaCl(aq) and NaOH(aq): NaCl is a salt and NaOH is a strong base. Thus, this system <em>is not </em> a buffer system.
H₂O(l) and HCl(aq): Water is a solvent and HCl a strong acid. This <em>is not </em>a buffer system.
HCl(aq) and NaOH(aq): HCl is a strong acid and NaOH a strong base. This <em>is not </em>a buffer system.
NaCl(aq) and NaNO₃(aq): Both NaCl and NaNO₃ are salts and this system <em>is not </em>a buffer system.
