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1 year ago

How many moles of CF4 can be produced when 8.95 mol C reacts with 7.88 mol F2?

Chemistry

1 answer:

Molodets [167]1 year ago

8 0

The remaining moles of C is 5.01 moles while the remaining moles of F₂ is 0.

<h3>Reaction between Carbon and Fluorine </h3>

The reaction between carbon and Fluorine is given as;

C + 2F₂ -------> CF₄

1 : 2 1

from the reaction above,

2 moles of F₂ requires 1 mole of C

7.88 mole of F₂ will require: 7.88/2 = 3.94 moles of C and 3.94 moles of CF₄.

The remaining moles of C = 8.95 - 3.94 = 5.01 moles while the remaining moles of F₂ is 0.

Learn more about moles here: brainly.com/question/15356425

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Which pair of elements is most likely to chemically combine and form ionic bonds, chlorine and chlorine or lithium and chlorine,

olya-2409 [2.1K]

I believe the correct answer from the choices listed above is the second option. The pair of elements that is most likely to chemically combine and form ionic bonds would be <span>lithium and chlorine. Lithium is metal and chlorine is nonmetal which as a compound forms ionic bonds. Hope this answers the question.</span>

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3 years ago

Consider a sample of 3.5 mol of N2(g) at T1 = 350 K, that undergoes a reversible and adiabatic change in pressure from p1 = 1.50

devlian [24]

Answer:

Part A is just T2 = 58.3 K

Part B ∆U = 10967.6 x C $_{V}$ You can work out C $_{V}$

Part C

Part D

Part E

Part F

Explanation:

P = n (RT/V)

V = (nR/P) T

P1V1 = P2V2

P1/T1 = P2/T2

V1/T1 = V2/T2

P = Pressure(atm)

n = Moles

T = Temperature(K)

V = Volume(L)

R = 8.314 Joule or 0.08206 L·atm·mol−1·K−1.

bar = 0.986923 atm

N = 14g/mol

N2 Molar Mass 28g

n = 3.5 mol N2

T1 = 350K

P1 = 1.5 bar = 1.4803845 atm

P2 = 0.25 bar = 0.24673075 atm

Heat Capacity at Constant Volume

Q = nCVΔT

Polyatomic gas: CV = 3R

P = n (RT/V)

0.986923 atm x 1.5 = 3.5 mol x ((0.08206 L atm mol -1 K-1 x 350 K) / V))

V = (nR/P) T

V = ((3.5 mol x 0.08206 L atm mol -1 K-1)/(1.5 x 0.986923 atm) )x 350K

V = (0.28721/1.4803845) x 350

V = 0.194 x 350

V = 67.9036 L

So V1 = 67.9036 L

P1V1 = P2V2

1.4803845 atm x 67.9036 L = 0.24673075 x V2

100.52343693 = 0.24673075 x V2

V2 = P1V1/P2

V2 = 100.52343693/0.24673075

V2 = 407.4216 L

P1/T1 = P2/T2

1.4803845 atm / 350 K = 0.24673075 atm / T2

0.00422967 = 0.24673075 /T2

T2 = 0.24673075/0.00422967

T2 = 58.3 K

∆U= nC $_{V}$ ∆T

Polyatomic gas: C $_{V}$ = 3R

∆U= nC $_{V}$ ∆T

∆U= 28g x C $_{V}$ x (350K - 58.3K)

∆U = 28C $_{V}$ x 291.7

∆U = 10967.6 x C $_{V}$

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