Answer:
6.43 moles of NF₃.
Explanation:
The balanced equation for the reaction is given below:
N₂ + 3F₂ —> 2NF₃
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Finally, we shall determine the number of mole of nitrogen trifluoride (NF₃) produced by the reaction of 9.65 moles of Fluorine gas (F₂). This can be obtained as follow:
From the balanced equation above,
3 moles of F₂ reacted to produce 2 moles of NF₃.
Therefore, 9.65 moles of F₂ will react to to produce = (9.65 × 2)/3 = 6.43 moles of NF₃.
Thus, 6.43 moles of NF₃ were obtained from the reaction.
Answer:
a) ΔGº= -49,9 KJ/mol = - 50 KJ/mol
b) The reaction goes to the right to formation of products
c) ΔG= 84,42 KJ/mol. The direction is to reactive, to the left
Explanation:
a) ΔGº= - RTLnKf
You need to convert Cº to K. 25ºC=298K
Then, ΔGº= - 3,814 J/molK * 298K* Ln(5.6 *10^8)= - 49906 J/mol = -49,9 KJ/mol = - 50 KJ/mol
b) The ΔGº < 0, that means the direct reaction is spontaneous when te reactive and products are in standard state. In other words the reaction goes to the right, to formation of products
c) The general ecuation for chemical reaction is aA + bB → cD + dD. Thus
ΔG=ΔGº + RTLn (([C]^c*[D]^d)/[A]^a*[B]^b)
In this case,
ΔG=ΔGº + RTLn ([Ni(NH3)62+] / [Ni2+]*[NH3]^6 )= 84417 J/mol =84,42 KJ/mol
ΔG >0 means the reaction isn't spontaneous in the direction of the products. Therefore the direction is to reactive, to the left
The reason why Br has a greater magnitude of electron affinity than that of I is that there is a greater attraction between an added electron and the nucleus in Br than in I.
In the periodic table, there are trends that increase down the group and across the period. Electron affinity is a trend that increases across the period but decreases down the group.
Recall that the ability of an atom to accept an electron depends on the size of the atom. The smaller the atom, the greater the attraction between an added electron and the nucleus.
Since Br is smaller than I, there is a greater attraction between an added electron and the nucleus in Br than in I which explains why Br has a greater magnitude of electron affinity than I.
Learn more: brainly.com/question/17696329
Answer:
The correct answer is A) Pb(C₂H₃O₂)₂ + Li₂SO₄
Explanation:
- All salts of Na, K and ammonium are soluble.
- All nitrates are soluble.
- All chlorides are soluble, with the exception of AgCl, Hg₂Cl₂, PbCl₂ and CuCl.
- All sulfates are soluble, with the exception of CaSO₄, SrSO₄, BaSO₄, PbSO₄, HgSO₄, Hg₂SO₄ and Ag₂SO₄.
- All hydroxides are poorly soluble, with the exception of alkaline hydroxides, Sr(OH)₂, Ba(OH)₂ and NH4(OH)
- All carbonates are poorly soluble, with the exception of alkaline carbonates and ammonium carbonate.
- All sulfides are poorly soluble, with the exception of alkaline sulfides, alkaline earth sulfides, and ammonium sulfide.
The reaction A is:
Pb(C₂H₃O₂)₂ + Li₂SO₄ ⇒PbSO₄ + 2 LiC₂H₃O₂
Lead sulfate is a slightly soluble solid, therefore it precipitates.
