If you have an aqueous solution that contains 1.5 moles of HCl, the number of moles of ions in the solution is 3.0 moles.
<h2>Further Explanation
</h2><h3>Strong acids </h3>
- Strong acids are types of acids that undergo complete dissociation to form ions when dissolved in water.
- Examples of such acids are, HCl, H2SO4 and HNO3
- Dissociation of HCl
HCl + H₂O ⇔ H₃O⁺ + OH⁻
<h3>Weak acids </h3>
- Weak acids are types of acids that undergo incomplete dissociation to form ions when dissolved in water.
- Examples of such acids are acetic acids and formic acids.
- Dissociation of acetic acid
H₃COOH ⇔ CH₃COO⁻ + H⁺; CH₃COO⁻ is a conjugate base of acetic acid.
<h3>In this case;</h3>
- HCl which is a strong acid that ionizes completely according to the equation;
HCl + H₂O ⇔ H₃O⁺ + OH⁻
- From the equation, 1 mole of HCl produces 1 mole of H₃O⁺ ions and 1 mole of OH⁻ ions.
Therefore;
1.5 moles of HCl will produce;
= 1.5 moles of H₃O⁺ ions and 1.5 moles of OH⁻ ions.
This gives a total number ions of;
= 1.5 + 1.5
= 3 moles of ions
Keywords: Strong acid, weak acid, ions, ionization
<h3>Learn more about: </h3>
Level: High school
Subject: Chemistry
Topic: Salts, Acids and Bases
Answer:
a women standing in high heels
Explanation:
HNO3+KOH = H2O+KNO3 . When nitric acid react with pottasuim hydroxide, the reaction will produce water (H20) and pottasuim trioxonitrate
Answer: Chosen landforms are:
1) Hill
2) Mountain
3) Plateau
4) Valley
Explanation:
1) Hill is an elevated location smaller than a mountain. Location: Land
2) Mountain is a large mass of earth and rock, rising above the common level of the earth or adjacent land, usually given by geographers as above 1000 feet in height (or 304.8 metres).
Location: Land or Water
3) Plateau is a largely level expanse of land at a high elevation. It is also known as tableland.
Location: Land
4) Valley is an elongated depression between hills or mountains, often with a river flowing through it.
Location: Land or Water
Answer:
Qm = -55.8Kj/mole
Explanation:
NaOH(aq) + HNO₃(aq) => NaNO₃(aq) + H₂O(l)
Qm = (mc∆T)water /moles acid
Given => 100ml(0.300M) NaOH(aq) + 100ml(0.300M)HNO₃(aq)
=> 0.03mole NaOH(aq) + 0.03mole HNO₃(aq)
=> 0.03mole NaNO₃(aq) + 0.03mole H₂O(l)
ΔH⁰rxn = [(200ml)(1.00cal/g∙°C)(37 – 35)°C]water / 0.03mole HNO₃
= 13,333 cal/mole x 4.184J/cal = 55,787J/mol = 55.8Kj/mole (exothermic)*
Heat of reactions comes from formation of H-Oxy bonds on formation of water of reaction and heats the 200ml of solvent water from 35⁰C to 37⁰C.
