Answer:
The answer to your question is V2 = 66.7 ml
Explanation:
Data
Volume 1 = V1 = 400 ml
Pressure 1 = P1 = 1 atm
Volume 2 = V2 = ?
Pressure 2 = P2 = 6 atm
Process
1.- To solve this problem use Boyle's law
P1V1 = P2V2
-solve for V2
V2 = P1V1 / P2
-Substitution
V2 = (1)(400) / 6
-Simplification
V2 = 400 / 6
-Result
V2 = 66.7 ml
Answer:
:0
Explanation:
One second after the Big Bang, the universe was filled with neutrons, protons, electrons, anti-electrons, photons and neutrinos.Jun 17, 2017
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
that's the answer I think
Answer:
for the given reaction is -99.4 J/K
Explanation:
Balanced reaction: 
![\Delta S^{0}=[1mol\times S^{0}(NH_{3})_{g}]-[\frac{1}{2}mol\times S^{0}(N_{2})_{g}]-[\frac{3}{2}mol\times S^{0}(H_{2})_{g}]](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20S%5E%7B0%7D%28NH_%7B3%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28N_%7B2%7D%29_%7Bg%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20S%5E%7B0%7D%28H_%7B2%7D%29_%7Bg%7D%5D)
where 
represents standard entropy.
Plug in all the standard entropy values from available literature in the above equation:
![\Delta S^{0}=[1mol\times 192.45\frac{J}{mol.K}]-[\frac{1}{2}mol\times 191.61\frac{J}{mol.K}]-[\frac{3}{2}mol\times 130.684\frac{J}{mol.K}]=-99.4J/K](https://tex.z-dn.net/?f=%5CDelta%20S%5E%7B0%7D%3D%5B1mol%5Ctimes%20192.45%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B1%7D%7B2%7Dmol%5Ctimes%20191.61%5Cfrac%7BJ%7D%7Bmol.K%7D%5D-%5B%5Cfrac%7B3%7D%7B2%7Dmol%5Ctimes%20130.684%5Cfrac%7BJ%7D%7Bmol.K%7D%5D%3D-99.4J%2FK)
So, for the given reaction is -99.4 J/K
