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KonstantinChe [14]
3 years ago
12

Which phrase describes the decay modes and the half-lives of K-37 and K-42?

Chemistry
2 answers:
True [87]3 years ago
7 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Among the choices the phrase describes the decay modes and the half-lives of K-37 and K-42 is different decay modes and different half-lives 
andrew11 [14]3 years ago
6 0

Answer: (3) different decay modes and different half-lives

Explanation: Potassium-37 and Potassium-42 have the same number of protons  but different numbers of neutrons i.e. they have same atomic number but different mass number. They are isotopes. Thus K-37 and K-42 have different decay modes

K-37 is positron decay and K-42 is beta decay.

_{19}^{37}{\textrm {Kr}}\rightarrow _{18}^{37}{\textrm {Ar}}+_{+1}^0{\textrm {e}}

_{19}^{42}{\textrm {Kr}}\rightarrow _{20}^{42}{\textrm {Ca}}+_{-1}^0{\textrm {e}}


K-37 has a half life of 1.23 seconds and K-42 has a half life of 12.4 hrs.

Thus they have different decay modes and different half lives.

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A 1.0 g sample of propane, C3H8, was burned in calorimeter. The temperature rose from 28.5 0C to 32.0 0C and heat of combustion
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Answer:

A 1.0 g sample of propane, C3H8, was burned in the calorimeter.

The temperature rose from 28.5 0C to 32.0 0C and the heat of combustion 10.5 kJ/g.

Calculate the heat capacity of the calorimeter apparatus in kJ/0C

Explanation:

Heat of combustion = heat capacity of calorimeter * deltaT\\

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The heat of combustion = 10.5kJ/g.

deltaT = (32.0-28.5)^oC\\deltaT = 3.5^oC

Substitute these values in the above formula to get the value of heat capacity of the calorimeter.

deltaT =heat  capacity of calorimeter   * (change in temperature)\\10.5kJ/g = heat  capacity of calorimeter * (3.5^oC)\\\\=>heat capacity of calorimeter = \frac{10.5kJ/g}{3.5^oC} \\=>heat capacity of calorimeter = 3.0 kJ/g.^oC

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5 0
3 years ago
A student increases the temperature of a 556 cm3 balloon from 278 K to 308 K. Assuming constant pressure, what should the new vo
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The answer is:  [D]:  " 417 cm³ " .
_____________________________________________________
Explanation:  Use the formula:

V₁ /T₁= V₂ /T₂  ;

in which:  V₁ = initial volume = 556 cm³ ;
                T₁ = initial temperature = 278 K ;
                V₂ = final ("new") temperature = 308 K
                T₂ = final ("new:) volume = ?

Solve for  "V₂" ;

Since:  V₁ /T₁= V₂ /T₂ ;

We can rearrange this "equation/formula" to isolate "V₂" on one side of the equation; and then we can plug in our know values to solve for "V₂" ;
_______________________________________________________
       V₁ /T₁= V₂ /T₂  ;  Multiply EACH side of the equation by "T₂ " :

          →  T₂ (V₁ /T₁) = T₂  (V₂ /T₂) ;
______________________________
to get:

↔ T₂  (V₂ /T₂) = T₂ (V₁ /T₁) ;

     →  V₂ = T₂ (V₁ /T₁) ;
______________________________
Now, plug in our known values, to solve for "V₂" ;
______________________________
    →  V₂ = T₂ (V₁ /T₁) ;
______________________________
    →  V₂ = 308 K ( 556 cm³ /278 K)  ;
             → The units of "K" cancel to "1" ; and we have:
________________________________________________________
    →  V₂ = 308*( 556 cm³ / 278 ) = [(208 * 556) / 278 ] cm³ ;
Note:  We will keep the units of volume as:  "cm³ ";  since all the answer choices given are in units of:  "cm³ " ; {that is, "cubic centimeters"}.

   →  [(208 * 556) / 278 ] cm³ = [ (115,648) / (278) ] cm³ ;
                                        
              → For the "(115,648)" ;  round to "3 (three significant figures)" ;
                        → "(115,648)" → rounds to:  "116,000" ;
____________________________________________________
              →      (116,000) / (278) = 417.2661870503597122  ;
                                 → round to 3 significant figures; → "417 cm³ " ;
                                               → which corresponds with "choice [D]".
______________________________________________________
The answer is:  [D]:  "417 cm³ " .
______________________________________________________

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