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Marizza181 [45]
3 years ago
13

What kind of nuclear reaction is seen in 236 92 U - 144 36 Kr 3p1n A; nuclear fission B; Beta decay C; Alpha decay D; Nuclear fu

sion
Chemistry
1 answer:
AlladinOne [14]3 years ago
6 0
<h3><u>Answer;</u></h3>

A; nuclear fission

<h3><u>Explanation</u>;</h3>
  • Nuclear fission is a process by which a large nucleus is split into two smaller nuclei, or fission fragments. It takes place after the nucleus absorbs a neutron that usually is a product of another atom's radioactive decay.
  • A common fission reaction bombards uranium-235 with a single neutron. This creates a uranium-236 atom which is unstable. Its nucleus tends to split into more stable products like Krypton-90 & Barium-143.
  • If you compare the masses before and after reaction, notice that Kr-90 + Ba-143 only = 233. But they were produced from U-236. This is because 3 neutrons were also released.
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3. Crystalline structural unit of barium metal is a body-centered cubic cell. The edge length of the unit cell is 5.02x10-8 cm.
tiny-mole [99]

Answer:

The Avogadro's  number is N_A     =  6.02289 *10^{23}

Explanation:

From the question we are told that

   The edge length is  L   = 5.02 * 10^{-8} \ cm= \frac{5.02 * 10^{-8} }{100}  =  5.02 * 10^{-10}

    The density of the metal is \rho =  5.30\ g/cm^3 = 5.30 * \frac{g}{cm^3}  * \frac{1*10^6}{1*10^3} = 5.30 *10^3kg/m^3

     The molar mass of  Ba is  Z  =  137.3 \ g/mol = \frac{137.3}{1000} =  0.1373 \  kg / mol

     

Generally the volume of a unit cell is  

       V =  L^3

substituting value

        V =  [5.02 *10^{-10}]^3

         V = 1.265*10^{-28}\ m^3  

From the question we are told that 68% of the unit cell is occupied by Ba atoms and that the structure is a metal which implies that the crystalline structure will be  (BCC),

The volume of barium atom is  

        V_a  =  \frac{V}{2}  * 0.68

substituting value

        V_a  =  \frac{ 1.265*10^{-28}}{2}  * 0.68

        V_a  = 4.301 *10^{-29} \ m^3

The Molar mass of barium is mathematically represented as

      Z  =  N_A V_a *  \rho

Where N_A is the Avogadro's number

 So  

      N_A     =   \frac{ Z}{ V_a *  \rho}

substituting value

     N_A     =   \frac{ 0.1373}{ 4.301*10^{-29} *  5.3*10^{3}}

     N_A     =  6.02289 *10^{23}

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Which of the following solutions would make a good buffer system? (Check all that apply.) A. A solution that is 0.10 M NH3 and 0
AlekseyPX

Answer:

A solution that is 0.10 M HCN and 0.10 M LiCN

. A solution that is 0.10 M NH3 and 0.10 M NH4Cl

Explanation:

A buffer consists of a weak acid and its conjugate base counterpart. HCN is a weak acid and the salt LiCN contains its counterpart conjugate base which is the cyanide ion. A buffer maintains the pH by guarding against changes in acidity or alkalinity of the solution.

A solution of ammonium chloride and ammonia will also act as a basic buffer. A buffer may also contain a weak base and its conjugate acid.

7 0
3 years ago
Read 2 more answers
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