Answer:6.719Litres of Cl2 gas.
Explanation:According to eqn of rxn
2Na +Cl2=2NaCl
P=689torr=689/760=0.91atm
T=39°C+273=312K
according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl
But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW
MW of NaCl=23+35.5=58.5g/mol
n=28g(mass given of NaCl)/58.5
n=0.479moles of NaCl
Going back to the reaction,
if 1moles of Cl2 produces 2moles of NaCl
x moles of Cl2 will give 0.479moles of NaCl.
x=0.479*1/2
x=0.239moles of Cl2.
To find the volume, we use ideal ggas eqn,PV=nRT
V=nRT/P
V=0.239*0.082*312/0.91
V=6.719Litres
84.34 grams of grams of iron (III) chloride that can be produced is maximum because Fe is the limiting reagent in this reaction and chlorine gas is excess reagent.
Explanation:
Balanced chemical equation:
2 Fe + 3 Cl2 → 2 FeCl3
DATA GIVEN:
iron = atoms
mass of chlorine gas = 67.2 liters
mass of FeCl3 = ?
number of moles of iron will be calculated as
number of moles = 
number of moles = 
number of moles = 0.52 moles of iron
moles of chlorine gas
number of moles = 
Putting the values in the equation:
n =
(atomic mass of chlorine gas = 70.96 grams/mole)
= 947.01 moles
Fe is the limiting reagent so
2 moles of Fe gives 2 moles of FeCl3
0.52 moles of Fe will give
= 
0.52 moles of FeCl3 is formed.
to convert it into grams:
mass = n X atomic mass
= 0.52 x 162.2 (atomic mass of FeCl3 is 162.2grams/mole)
<h3> = 84.34 grams </h3>
Answer:
37.65mL
Explanation:
Given parameters:
density of liquid Z = 0.9237g/mL
Mass of liquidZ + mass of cup = 50.7g
Mass of cup= 15.92g
Volume of liquid Z in cup=?
Solution:
Density is the mass per unit volume of a substance. It is mathematically expressed as shown below:
Density = 
To find the volume of liquid Z, we know the density of the liquid but we dont know the mass yet.
Mass of liquidZ = 50.7g - mass of cup = 50.7g - 15.92g = 34.98g
Therefore:
Volume of liquidZ = 
=
= 37.65mL
It would be elliptical according to him.....
Given reaction represents dissociation of bromine gas to form bromine atoms
Br2(g) ↔ 2Br(g)
The enthalpy of the above reaction is given as:
ΔH = ∑n(products)Δ
- ∑n(reactants)Δ
where n = number of moles
Δ
= enthalpy of formation
ΔH = [2*ΔH(Br(g)) - ΔH(Br2(g))] = 2*111.9 - 30.9 = 192.9 kJ/mol
Thus, enthalpy of dissociation is the bond energy of Br-Br = 192.9 kJ/mol