The sample is most likely lead with lighter substances in it. Pure lead would be closer to the pure lead density, and heavier substances would make the density more, not less.
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
<span>08 moles Li3N * 1mole N2/2moles Li3N = 0.04 </span>
Better than i am and very precice
Answer:
Molarity of NaOH = 1.8 M.
Explanation:
From the question given above, the following data were obtained:
Mass of NaOH = 36 g
Molar mass of NaOH = 40 g/mol
Volume = 500 mL
Molarity of NaOH =?
Next, we shall determine the number of mole in 36 g of NaOH. This can be obtained as follow:
Mass of NaOH = 36 g
Molar mass of NaOH = 40 g/mol
Mole of NaOH =?
Mole = mass / molar mass
Mole of NaOH = 36 / 40
Mole of NaOH = 0.9 mole
Next, we shall convert 500 mL to L. This can be obtained as follow:
1000 mL = 1 L
Therefore,
500 mL = 500 mL × 1 L / 1000 mL
500 mL = 0.5 L
Finally, we shall determine the molarity of NaOH. This can be obtained as follow:
Mole of NaOH = 0.9 mole
Volume = 0.5 L
Molarity of NaOH =?
Molarity = mole / Volume
Molarity of NaOH = 0.9 / 0.5
Molarity of NaOH = 1.8 M
