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3 years ago

I need Coefficient numbers!!!!!

Chemistry

1 answer:

KengaRu [80]3 years ago

4 0

Not sure about #6 , but i got all the other answers

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The 225 g FeCl2 is about 1.8 moles.

Pachacha [2.7K]

Answer:

I just guessed it out to be 3

8 0

1 year ago

A garbage collector collects food wrappers and aluminum cans from a park. What type of waste is the garbage collector collecting

Nat2105 [25]

Your answer to this question would be B. packaging waste.

hope this helps!

3 0

2 years ago

Read 2 more answers

One mole of an ideal gas, for which CV,m = 3/2R, initially at 298 K and 1.00 × 105 Pa undergoes a reversible adiabatic compressi

oksian1 [2.3K]

Answer:

final temperature (T2) = 748.66 K
ΔU = w = 5620.26 J
ΔH = 9367.047 J
q = 0

Explanation:

ideal gas:

PV = RTn

reversible adiabatic compression:

δU = δq + δw = CvδT

∴ q = 0

∴ w = - PδV

⇒ δU = δw

⇒ CvδT = - PδV

ideal gas:

⇒ PδV + VδP = RδT

⇒ PδV = RδT - VδP = - CvδT

⇒ RδT - RTn/PδP = - CvδT

⇒ (R + Cv,m)∫δT/T = R∫δP/P

⇒ [(R + Cv,m)/R] Ln (T2/T1) = Ln (P2/P1) = Ln (1 E6/1 E5) = 2.303

∴ (R + Cv,m)/R = (R + (3/2)R)/R = 5/2R/R = 2.5

⇒ Ln(T2/T1) = 2.303 / 2.5 = 0.9212

⇒ T2/T1 = 2.512

∴ T1 = 298 K

⇒ T2 = (298 K)×(2.512)

⇒ T2 = 748.66 K

⇒ ΔU = Cv,mΔT

⇒ ΔU = (3/2)R(748.66 - 298)

∴ R = 8.314 J/K.mol

⇒ ΔU = 5620.26 J

⇒ w = 5620.26 J

H = U + nRT

⇒ ΔH = ΔU + nRΔT

⇒ ΔH = 5620.26 J + (1 mol)(8.314 J/K.mol)(450.66 K)

⇒ ΔH = 5620.26 J + 3746.787 J

⇒ ΔH = 9367.047 J

8 0

3 years ago

The strongest forces of attraction occur between molecules of(1) HCl (3) HBr(2) HF (4) HI

Alexxandr [17]

It would likely be HF. It displays hydrogen bonding.

4 0

3 years ago

1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride ? atoms of boron.

maks197457 [2]

Answer:

1. How many ATOMS of boron are present in 2.20 moles of boron trifluoride? atoms of boron.

2. How many MOLES of fluorine are present in of boron trifluoride? moles of fluorine.

Explanation:

The molecular formula of boron trifluoride is $BF_3$ .

So, one mole of boron trifluoride has one mole of boron atoms.

1. The number of boron atoms in 2.20 moles of boron trifluoride is 2.20 moles.

The number of atoms in 2.20 moles of boron is:

One mole of boron has ---- $6.023x10^2^3$ atoms.

Then, 2.20 moles of boron has

- $=2.20 mol. x 6.023 x 10^2^3 atoms /1 mol\\=13.25x10^2^3 atoms$

2. Calculate the number of moles of BF3 in 5.35*1022 molecules.

$(5.35x10^2^2 molecules/6.023x10^2^3)x 1mol\\=0.0888mol$

One mole of boron trifluoride has three moles of fluorine atoms.

Hence, 0.0888moles of BF3 has 3x0.0888mol of fluorine atoms.

=0.266mol of fluorine atoms.

5 0

2 years ago