Question

5–9 pleae show work

5. A gas sample at 1.0 atm and 25°C is heated at constant pressure until its volume is doubled. What is the final temperature of the gas? a. 298°C b. 323°C c. 596°C d. 50°C e. None of the above 6. The bond order of C-O bond in Co, is b. 1.33 c. 1.5 d. 2.0 e. None of the above 7. The electron configuration of copper ion (Cu) is a. [Ar] 4s 3d b. [Ar] 4s 310 c. [Ar] 4s 3d d. [Ar]3d e. None of the above 8. What is the molarity of concentrated sulfuric has the density of 1.84 g/mL? 8% (by mass) and a. 0.056 mol/L b. 0.56 mol/L c. 9.00 mol/L d. 18.0 mol/L e. None of the above 9. P. + F → PF, has a yield of 87% g/mol) gas are produced from 25 g of F. (molar mass = 38 g/mol) gas? How many grams of PF, (molar mass - 88 a. 33.6 g b. 44.48 c. 77.6 g d. 174.65 e. None of the above

Answer 1

5) use ideal gas relation to solve this
At constant pressure, V1/T1 = V2/T2

let initial volume V1 = x
Then final volume V2 = 2x [since volume is doubled]
Initial temperature T1= 25 C = 298 K
Then final Temperature T2= ???
T2 = V2*T1/V1
= 2x *298/x = 596 K
= [596 -273] C = 323 C

hence answer option B
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6)
There are two double bonds and 2 single bonds in oxalate ion    in the lewis structure
number of bond groups/total number of bonds = bond order
So bond order = [1+2+1+4/]4 = 1.5

Answer option C
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7)
Atomic number for Cu= 29
electron configuration = [Ar] 4s1 3d10
Cu+ ion means 1 electron less than Cu atom.This one electron will remove from
4s subshell,since it isthe outer shell

electron configuration of Cu+ = [Ar] 3d10

Option D
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8)
mass percent means mass of solute/mass of solution x100

48 % H2SO4 means 48 g of H2SO4 in 100 g of solution

convert this mass in to moles of H2So4
moles of H2SO4 = mass of H2So4/molar mass = 48 g/98 g/mol
=0.48979 mols**
Now convert mass of solution to volume using given density value
volume of solution = mass /density = 100 g/1.84 g/ml = 54.347 ml
= 0.054347 liters**
now
molarity= moles of H2SO4/volume in L =0.48979 mols/0.054347 liters
= 9 mols/L
option C
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9) write balanced equation
P4 +6 F2 ---> 4PF3
mole ratio of F2:PF3 = 6:4

moles of F2 present = massof F2/molar mass of F2 =25g/38 g/mol =0.658 mols
SO moles of PF3 formed = {4/6}x0.658 =0.439 moles
mass of PF3 produced = moles xmolar mass = 0.439 mole x88 g/mol=38.632g

But actual yield is 87 %
SO actual grams of PF3 =38.596 x87/100= 33.6 g
=

Option A
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