The change in internal energy of the combustion of biphenyl in Kj is calculated as follows
=heat capacity of bomb calorimeter x delta T where delta T is change in temperature
delta T = 29.4 -25.8= 3.6 c
= 5.86 kj/c x 3.6 c = 21.096 kj
Answer:
F2 is the limiting reactant
27.6 grams of NaF is produced.
Explanation:
Balance the equation first.
2Na+ F2 ---> 2NaF
To find the limiting reactant, solve for how much NaF can be produced with Na and F2
12.5g F2 x (1 mole F2/ 38.00 grams F2)x (2 mole NaF/ 1 mole F2)
=0.658 moles NaF
16.2g Na x (1 mole Na/ 22.99 grams Na)x (2 mole NaF/ 2 mole Na)
=0.705 moles NaF
Since F2 produced the least NaF, F2 is the limiting reactant.
Now, to find how much NaF there is, use the moles solved above with F2 as the limiting reactant.
0.658 moles NaF x (41.99 grams NaF/ 1 mole NaF)= 27.6 moles NaF
27.6 moles of NaF would be theoretically produced.
Ionization energy is the measure of the extend to which the nucleus attracts the outermost electron
if ionization energy us high than force of attraction Is high so it is not easy to remove and vice versa .
hope you understand.....
An aqueous solution contains the following ions Cl⁻, Ag⁺, Pb²⁺, NO₃⁻ & SO₄²⁻ and more than one precipitate will form are AgCl, PbCl₂, PbSO₄ & Ag₂SO₄.
<h3>What is precipitate?</h3>
Precipitate is the insoluble compound which is present at the bottom of any chemical reaction in the solid state.
If in an aqueous solution Cl⁻, Ag⁺, Pb²⁺, NO₃⁻ & SO₄²⁻ ions are present then:
- Compounds AgCl, PbCl₂, PbSO₄ & Ag₂SO₄ are not soluble in water as it is present in the form of precipitate.
- Pb(NO₃)₂ is fully soluble in water and will not make precipitate.
Hence precipitates are AgCl, PbCl₂, PbSO₄ & Ag₂SO₄.
To know more about precipitates, visit the below link:
brainly.com/question/2437408
#SPJ4
