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kupik [55]
3 years ago
7

Which descriptions apply to emitted radiation? Check all that apply. measurement in Ci/Bq measurement in conventional rad or the

SI Gy the amount of radioactive materials released into the environment. number of disintegrations of radioactive atoms in a radioactive material over a period of time absorbed dose (the amount of energy deposited per unit of mass of human tissue)
Chemistry
2 answers:
xxTIMURxx [149]3 years ago
7 0
The correct answer is measurement in Ci/Bq, the amount of radioactive materials released into the environment, and <span>number of disintegrations of radioactive atoms in a radioactive material over a period of time. I hope this helps.</span>
Llana [10]3 years ago
6 0

Answer:

- Measurement in Ci/Bq

- The amount of radioactive materials released into the environment.

- Number of disintegrations of radioactive atoms in a radioactive material over a period of time.

Explanation:

Hello,

In this case, the following options are valid:

- Measurement in Ci/Bq : one becquerel is understood as the quantity of a radioactive element in which there is one atomic disintegration per second.

- The amount of radioactive materials released into the environment: this technique is suitable to compute the amount of radioactive materials that are being or were released to the environment.

- Number of disintegrations of radioactive atoms in a radioactive material over a period of time: this technique provides becquerel which were described on the first option.

Best regards.

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7. A 2.0 L container had 0.40 mol of He(g) and 0.60 mol of Ar(g) at 25°C.
Finger [1]

Answer:

a) Ek Ar > Ek He

b) v Ar < v He

c) If v Ar = 431 m/s ⇒ v He = 1710.44 m/s

d) Pt = 12.218 atm

e) P He = 4.887 atm and P Ar = 7.33 atm

Explanation:

container:

∴ V = 2.0 L

∴ n He = 0.4 mol

∴ n Ar = 0.6 mol

∴ T = 25°C ≅ 298 K

a) Internal energy (U) :

∴ U = Ek + Ep = kinetic energy + potential energy

∴ Ep: the potential interaction energy is neglected, assuming ideal gas mixture

⇒ U = Ek = N(1/2mv²)= 3/2 NKT

∴ N = nNo ....number of moleculas

∴ K = 1.380 E-23 J/K....Boltzmann's constant

∴ No = 6.022 E23 molec/mol....Avogadro's number

for He:

⇒ N = (0.4)(6.022 E23) = 2.4088 E23 molec

⇒ Ek = (3/2)(2.4088 E23)(1.380 E-23 J/K)(298) = 1485.892 J

for Ar:

⇒ N = (0.6)(6.022 E23) = 3.6132 E 23 molec

⇒ Ek = (3/2)(3.6132 E23)(1.380 E-23 J/K)(298) = 2228.838 J

** Ar gas has a greater average kinetic energy

b) He:

∴ N(1/2)mv² = (3/2)NKT

⇒ mv² = 3KT

⇒ v² = 3KT/m

⇒ v = √3KT/m

∴ m He = (0.4 mol)(4.0026 g/mol) = 1.601 g He = 1.601 E-3 Kg He

⇒ v = √(3(1.380 E-23)(298)/(1.601 E-3)) = 2.776 E-9 m/s He

Ar:

∴ m Ar = (0.6)(39.948 g/mol) = 23.969 g = 0.0239 Kg Ar

⇒ v = 6.99 E-10 m/s

** v Ar < v He

c) r = V Ar / v He = (6.99 E-10 m/s)/(2.776 E-9 m/s) = 0.252

∴ If v Ar = 431 m/s

⇒ v He = v Ar/0.252 = 431 m/s / 0.252 = 1710.44 m/s

d) Pt = ntRT / V

∴ nt = 0.4 + 0.6 = 1 mol

⇒ Pt = (1mol)(0.082 atm.L/K.mol)(298 K)/(2.00 L) = 12.218 atm

e) P He = nRT/V = (0.4)(0.082)(298)/2 = 4.8872 atm

⇒ P Ar = Pt - PHe = 12.218 - 4.8872 = 7.33 atm

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