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Nesterboy [21]
3 years ago
12

Which of the following elements is stable?

Chemistry
1 answer:
larisa86 [58]3 years ago
3 0
It is neon due to its configuration 2,8. it has an octet structure so it is stable and inert.
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What is the mass of 1.25 mol of zinc ​
Ilya [14]

Answer:

1 mole is equal to 1 moles Zinc, or 65.38 grams.

Explanation:

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What is the name of the molecule shown below?
xxTIMURxx [149]

Answer:

B. 1-heptene

Explanation:

The 1-HEPTENE molecule contains a total of 20 bond(s) There are 6 non-H bond(s), 1 multiple bond(s), 4 rotatable bond(s) and 1 double bond(s).

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2 years ago
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2NO (g) + O2 (g) →2NO2 (g) At equilibrium [NO] = 2.4 × 10 -3 M, [O2] = 1.4 × 10 -4 M, and [NO2] = 0.95 M.
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Answer:

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Explanation:

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Unfortunately, the question is not given in the question; however, it is possible for us to compute the equilibrium constant as the problem is providing the concentrations at equilibrium. Thus, we first set up the equilibrium expression as products/reactants:

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K=\frac{(0.95)^2}{(0.0024)^2(0.00014)}\\\\K=1.12x10^9

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4 0
3 years ago
What determines the state of a substance
frez [133]

The kinetic energies of the particles (atoms, molecules, or ions) that make up a substance or object.

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3 years ago
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If 0.500 mol neon at 1.00 atm and 273 K expands against a constant external pressure of 0.100 atm until the gas pressure reaches
trapecia [35]

Answer:

The work done on neon = -323 J

The internal energy change= -392.84 J

The heat absorbed by neon = -69.84 J

Explanation:

Step 1: Data given

Number of moles  = 0.500 moles

Pressure  = 1 atm

Temperature  = 273 Kelvin

The pressure will change from 1.00 atm to 0.200 atm. The temperature changes from 273 to 210 Kelvin.

a) calculate the work done on neon

W = -P(V2-V1)    

⇒ with P = the pressure = 0.1 atm

⇒ with V1 = the initial volume = nRTi /Pi

⇒ with V2 = the final volume = nRTf /Pf

W = -PnR((T2/P2) -(T1/P1))

⇒ with T2 = the final temperature = 210 K

 ⇒ with T1 = the initial temperature = 273 K

 ⇒ with P2 = the final pressure = 0.200 atm

 ⇒ with P1 = the initial pressure = 1.00 atm

W = -nR (210*(0.1/0.2) - 273*(0.1/1.00))

W = -nR*(105 - 27.3)

W= -(0.500)*(8.314)*(77.7)

W = -323 J

b) calculate the internal energy change

E = (3/2)*nRT

ΔE = Ef - Ei

ΔE =(3/2)*nR(T2-T1)

⇒ with n= number of moles = 0.500 moles

⇒ with T2 =the final temperature = 210 K

⇒ with T1 = the initial temperature = 273 K

ΔE = (3/2)*(0.5)*(8.314)(210-273)

ΔE = -392.84 J

c) Calculate the heat absorbed by neon

ΔE = q + W

q = ΔE -W

⇒ with ΔE = -392.84 J

⇒ with W = -323 J

q = -392.84 J -( -323 J)

q =-392.84 J + 323 J

q = -69.84 J

4 0
4 years ago
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