He could be blindfolded and know which was his and which was his sister's. All he would need to do is pick them both up and if they were too big then pick them up one at a time. The lumber might make it harder to tell, but this is a question about physical properties.
So there is a change in mass which for the purpose of this question should be quite different. His sister's ought to be much lighter than his. He would find it easier to pick up.
The kind of reaction that occurs when you mix aqueous solutions of barium sulfide and sulfuric acid is a precipitation reaction.
<h3>Further Explanation</h3>
- The chemical reaction between Ba(OH)2(aq) and H2SO4(aq) is given by;
Ba(OH)₂(aq) + H₂SO4(aq) --> BaSO₄(aq) + 2H₂O(l)
- This is a type of precipitation reaction, where a precipitate is formed after the reaction, that is Barium sulfate.
<h3>Other types of reaction</h3><h3>Neutralization reactions </h3>
- These are reactions that involve reacting acids and bases or alkali to form salt and water as the only products.
- For example a reaction between sodium hydroxide and sulfuric acid.
NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + H₂O(l)
<h3>Displacement reactions</h3>
- These are reactions in which a more reactive atom or ion displaces a less reactive ion from its salt.
Mg(s) + CuSO₄(aq) → MgSO₄(aq) + Cu(s)
<h3>Redox reactions </h3>
- These are reactions that involve both reduction and oxidation occuring simultaneously durin a chemical reaction.
- For example,
Mg(s) + CuSO₄(aq) → MgSO₄(aq) + Cu(s)
- Magnesium atom undergoes oxidation while copper ions undergoes reduction.
<h3>Decomposition reactions</h3>
- These are type of reactions that involves breakdown of a compound into its constituents elements.
- For example decomposition of lead nitrate.
Pb(NO3)2(S) → PbO(s) + O2(g) + NO2(g)
Keywords: Precipitation
<h3>Learn more about: </h3>
Level: High school
Subject: Chemistry
Topic: Chemical reactions
Sub-topic: Precipitation reactions
Answer:
So she can have something to reach or look forward to.
Explanation:
none
Answer:
Molecules move from areas of high concentration to areas of low concentration.
Answer:
300000Pa or 3×10^5 Pa
Explanation:
Since the problem involves only two parameters of volume and pressure, the formula for Boyle's law is suitably used.
Using Boyle's law
P1V1 = P2V2
P1 is the initial pressure = 1.5×10^5Pa
V1 is the initial volume = 0.08m3
P2 is the final pressure (required)
V2 is the final volume = 0.04 m3
From the formula, P2 = P1V1/V2
P2 = 1.5×10^5 × 0.08 ÷ 0.04
= 300000Pa or 3×10^5 Pa.
