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AleksandrR [38]
3 years ago
12

A sample of pure NO2 is heated to 338 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+

O2(g) At equilibrium the density of the gas mixture is 0.515 g/L at 0.745 atm .
Chemistry
1 answer:
natima [27]3 years ago
3 0

Answer:

A sample of pure NO2 is heated to 338 ∘C at which temperature it partially dissociates according to the equation 2NO2(g)⇌2NO(g)+O2(g) At equilibrium the density of the gas mixture is 0.515 g/L at 0.745 atm .

(4x^2)x

Kc= -----------

(A-2x)^2

PV=nRT

n/v = P/RT = .745/(0.0821)(334+273) = .01495

To Find the initial molarity of NO2

(mol/L)(g/mol) + (mol/L)(g/mol) + (mol/L)(g/mol)= g/L

Thus:

46(A-2x) + 2x(30) + 32x = .515 g/L

46A-92x+60x+32x = .515

46A=.515

A=.01120 M

Using the total molarity found

(A-2x)+2x+x = .01495 M

A+x=.01495

Plug in A found into the above equation:

.01120+x = .01495

x=.00375

Now Plug A and x into the original Equilibrium Constant Expression:

(4x^2)x

Kc= -----------

(A-2x)^2

Kc = 0.000014

Explanation:

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Iodine-131 is a radioactive isotope that is used in the treatment of cancer of the thyroid. The natural tendency of the thyroid
Finger [1]

Answer:

a) Binding Energy = 1084.266 MeV

b) Binding energy per nucleon = 8.28 MeV

Explanation:

Binding Energy, BE = \triangle M * 931.5

Where ΔM = Mass defect of Iodine-131

The mass of iodine-131 = 130.906124 u.

Number of protons in Iodine-131 = 53

Number of neutrons in Iodine-131 = 78

Mass of a proton = 1.007276 u

Mass of neutron = 1.008664 u

Total mass of the 53 protons in Iodine-131 = (53*1.007276)

Total mass of the 53 protons in Iodine-131 =  53.39 u

Total mass of the 78 protons in Iodine-131 = (78*1.008664)

Total mass of the 78 protons in Iodine-131 = 78.68 u

\triangle M = (53.39 + 78.68) - 130.906124\\\triangle M = 1.164 u

Binding Energy = 1.164 * 931.5

Binding Energy = 1084.266 MeV

b) Binding Energy per nucleon

The mass of the nucleon of iodine 131 = 130.906124

Binding Energy per nucleon =  1084.266/130.906124

Binding energy per nucleon = 8.28 MeV

5 0
4 years ago
How many grams of solid Na2CO3 are required to neutralize exactly 2 liters of an HCI solution of pH 2.0?
Snezhnost [94]

Answer:

The answer is 1.06g.

Explanation:

Analysis of question:

1. Identify the information in the question given.

  • volume of HCl is 2 dm3
  • pH of HCl is 2.0

2. What the question want?

  • mass of Na2CO3 is ?(unknown)
  • 3. Do calculation.
  • 1st-Write a balanced chemical equation:

Na2CO3 + 2HCl (arrow) 2NaCl + H20 + CO2

  • 2nd-Determine the molarity of HCl with the value of 2.0.

pH= -log[H+]

2.0= -log[H+]

log[H+]= -2.0

[H+]= 10 to the power of negative 2(10-2)

=0.01 mol dm-3

molarity of HCl is 0.01 mol dm-3

  • 3rd-Find the number of moles of HCl

n=MV

=0.01 mol dm-3 × 2 dm3

=0.02 mol of HCl

  • 4th-Find the second mol of it.

Based on the chemical equation,

2.0 mol of HCl reacts with 1.0 mol of Na2CO3

0.02 mol of HCl reacts with 0.01 mol of Na2CO3

<u>N</u><u>a</u>2CO3>a=<u>1</u><u> </u>mol

<u>2</u><u>H</u>Cl>b=<u>2</u><u> </u>mol

  • 5th-Find the mass of it.

mass= number of mole × molar mass

g=0.01 × [2(23)+ 12+ 3(16)]

g=0.01 × 106

# =1.06 g.

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