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Tema [17]
3 years ago
12

How many joules are released when one atom of americium 241, the isotope used in ionization-type smoke detectors, undergoes apha

emission?
Chemistry
1 answer:
klemol [59]3 years ago
4 0

When Americium (Am-241) undergoes alpha decay(He-4) it forms neptunium (Np-237) based on the following pathway:

²⁴¹Am₉₅ → ²³⁷Np₉₃ + ⁴He₂

The energy released in given as:

ΔE = Δmc²

where Δm = mass of products - mass of reactants

                 = [m(Np-237) + m(He-4)] - [m(Am-241)]

                 = 237.0482+4.0015-241.0568 = -0.0073 g/mol = -7.3 * 10⁻⁶ kg/mol

ΔE = -7.3*10⁻⁶ kg/mol * (3*10⁸ m/s)² = -5.84*10¹¹ J/mol



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What is the vapor pressure of an aqueous solution made up of 60.0g of urea (CH4N2O) in 180g of water? the vapor pressure of wate
djyliett [7]
To get the answer you use the Law of Raoult.


Raoult's law states that the decrease of the vapor pressure of a liquid is proportional to the molar fraction of the solute.


ΔP = Pa * Xa


Here Pa = 0.038 atm


And Xa = N a / (Na + Nb), where Na is number of moles of A and Nb is number of moles of b


Na = mass of urea / molar mass of urea =  60 g / (molar mass of CH4N2O)


molar mass of CH4N2O = 12 g/mol + 4*1g/mol + 2*14 g/mol + 16 g/mol = 60 g/mol


Na = 60 g / 60 g/mol = 1 mol


Nb = mass of water / molar mass of water = 180g / 18g/mol = 10 mol


Xa = 1 mol / (10 mol + 1 mol) = 1/11 =0.09091


ΔP = Pb * Xa = 0.038 atm * 0.09091 =  0.0035 atm


Then, the final vapor pressure of water is Pb - ΔP = 0.038atm - 0.0035atm = 0.035 atm.


 Answer: 0.035 atm
7 0
3 years ago
Suppose that a catalyst lowers the activation barrier of a reaction from 125kJ/mol to 55kJ/mol125⁢kJ/mol to 55kJ/mol. By what fa
sineoko [7]

Answer:

The factor of increasing reaction rate is 1,85x10¹².

Explanation:

Using arrhenius formula:

k = A e^\frac{-E_{a}}{RT}

Where k is rate constant; A is frecuency factor; Eₐ is activation energy; R is gas constant (0,008134 kJ/molK); T is temperature 25°C = 298,15K

Thus, replacing for an activation energy of 125 kJ/mol assuming A as 1:

k = 1,25x10⁻²²

When activation energy is 55kJ/mol:

k = 2,31x10⁻¹⁰

Thus, the factor of increasing reaction rate is:

2,31x10⁻¹⁰/1,25x10⁻²² =<em> 1,85x10¹²</em>

<em></em>

I hope it helps!

8 0
3 years ago
Which type of hybridization is observed when central atom is located at the centre of regular tetrahedron
Ann [662]

Answer:

sp3 hybridization

Explanation:

Hybridization means the mixing of atomic orbitals to yield hybrid orbitals with characteristics that are different from that of the isolated atomic orbitals before the combination.

sp3 hybridization occurs when one s orbital is mixed with three p orbitals to yield four sp3 hybrid orbitals which can be used to bond to a central atom.

The central atom is then located at the center of a regular tetrahedron at a bond angle of 109°.

3 0
2 years ago
Indicate the correct number of sig figs
Dimas [21]

Explanation:

The answer to questions are

A) 4

B) 3

C) 5

D) 3

E) 3

3 0
3 years ago
Calculate the maximum wavelength of light that will cause the photoelectric
Drupady [299]

Answer:

Explanation:

Work function of potassium = 2.29 eV = 3.67 X 10⁻¹⁹ J

So the minimum energy of photon must be equal to 3.67 X 10⁻¹⁹ J .

energy of photon of wavelength λ = hc / λ

where h = 6.67  x 10⁻³⁴

c = 3 x 10⁸

Putting the values in the equation above

6.67  x 10⁻³⁴  x  3 x 10⁸ / λ =  3.67 X 10⁻¹⁹

λ  = 6.67  x 10⁻³⁴  x  3 x 10⁸ /  3.67 X 10⁻¹⁹

= 5.452 x 10⁻⁷

= 5452 x 10⁻¹⁰ m

= 5452 A .

6 0
3 years ago
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