Mass is a basic property of matter and is made of atoms. Atoms are made of electrons, protons, and neutrons. Protons and neutrons make up the majority of the mass in an atom. Atoms may vary in the number of neutrons they have (called isotopes), and therefore the exact atomic mass may vary from atom to atom
Molecules arranged in regular pattern changes into an irregular pattern
Answer:
2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)
Explanation:
The net ionic equation usually shows the main ionic reaction that goes in the system. The other ions that do not participate in this net ionic equation are called spectator ions. Spectator ions do not participate in the main reaction occurring in the system.
The net ionic equation quite often result in the formation of a solid precipitate in the system such as Cu(OH)2.
The net ionic equation for this reaction is;
2OH^-(aq) + Cu^2+(aq) -----> Cu(OH)2(s)
You can use P1V1/T1 = P2V2/T2 but since pressure is constant is becomes V1/T1=V2/T2
V1=0.5 L
T1=203 K
T2=273 K
V2=unknown
0.5L/203 = V2/273
V2= 0.67 L so C
Hope this helps :)
Answer:
a)
,
, b)
, 
Explanation:
a) The ideal gas is experimenting an isocoric process and the following relationship is used:

Final temperature is cleared from this expression:


The number of moles of the ideal gas is:



The final temperature is:


The final pressure is:



b) The ideal gas is experimenting an isobaric process and the following relationship is used:

Final temperature is cleared from this expression:




The final volume is:


