<u>Given:</u>
Initial volume of He, V1 = 19.2 L
Initial mass of He, m1 = 0.0860 g
Mass of He removed = 0.205 g
<u>To determine:</u>
The new volume of He i.e V2
<u>Explanation:</u>
Based on Avogadro's law:
Volume of a gas is directly proportional to the # moles of the gas
Volume (V) α moles (n) -----(1)
Atomic mass of He = 4 g/mol
Initial moles of He, n1 = 0.860 g/4 g.mol-1 = 0.215 moles
Final moles of He, n2 = (0.860-0.205)g/4 g.mol-1 = 0.164 moles
Based on eq(1) we have:
V1/V2 = n1/n2
V2 = V1 n2/n1 = 19.2 L * 0.164 moles/0,215 moles = 14.6 L
Ans: New volume is 14.6 L
Answer:
Since with LiBr no precipitation takes place. So, Ag+ is absent
When we add Li2SO4 to it, precipitation takes place.
Ca2+(aq) + SO42-(aq) ----> CaSO4(s) ...Precipitate
Thus, Ca2+ is present.
When Li3PO4 is added, again precipitation takes place.Reaction is:
Co2+(aq) + PO43-(aq)---->Co3(PO4)2(s) ... Precipitate
A. Ca2+ and Co2+ are present in solution
B. Ca2+(aq) + SO42-(aq) ----> CaSO4(s)
C. 3Co2+(aq) + 2PO43-(aq)---->Co3(PO4)2(s)
To be able to answer this equations, we must set given information. Suppose the reaction to yield NO is:
N₂ + O₂ → 2 NO
Next, suppose you have 1 g of each of the reactants. Determine first which is the limiting reactant.
1 g N₂ (1 mol N₂/ 28 g)(2 mol NO/1 mol N₂)= 0.07154 mol NO present
Number of molecules = 0.07154 mol NO(6.022×10²³ molecules/mol)
<em>Number of molecules = 4.3×10²² molecules NO present</em>
Answer: electrons
Explanation: moving electrons cause momentarily charge
Distribution on molecule. This distribution induces similar distribution to
Adjacent molecule.
