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Rudiy27
3 years ago
11

A decomposition reaction has a half life of 578 years. (a) What is the rate constant for this reaction? (b) How many years does

it take for the reactant concentration to reach 12.5% of its original value? Report your answer to 3 significant figures.
Chemistry
1 answer:
maw [93]3 years ago
3 0

Answer:

0.001199 year^{-1} is the rate constant for this reaction.

It will take 1.73\times 10^3 years to concentration to reach 12.5% of its original value.

Explanation:

A decomposition reaction follows first order kinetics:

Half life of the reaction = t_{1/2}=578 years

Rate constant of the reaction = k

For first order reaction, half life and rate constant are linked with an expression :

k=\frac{0.693}{t_{1/2}}

k=\frac{0.693}{578 years}=0.001199 year^{-1}

0.001199 year^{-1} is the rate constant for this reaction.

Initial concentration of reactant =[A_o] =  x

Final concentration of reactant after time t =[A] =  12.5% of x = 0.125x

The integrated law of first order reaction :

[A]=[A_o]\times e^{-kt}

0.125x=x\times e^{-0.001199 year^{-1}\times t}

t = 1,734.31 years =1.73\times 10^3 years

It will take 1.73\times 10^3 years to concentration to reach 12.5% of its original value.

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Please help with 3! Please give only the correct answer...
cupoosta [38]
The answer is:  " 1.75 * 10 ^(-10)  m " .
_________________________________________________________
Explanation: 
_________________________________________________________
This very question asked for "Question Number 3 (THREE) ONLY, which is fine!
_________________________________________________________
Given: " 0.000000000175 m " ;  write this in "scientific notation.
_________________________________________________________
Note:   After the "first zero and the decimal point" {Note: that first zero that PRECEDES the decimal point in merely a "placeholder" and does not count as a "digit" — for our purposes} —
                     There are NINE (9) zeros, followed by "175"
_______________________________________________________
To write in "scientific notation", we find the integer that is written, as well, as any "trailing zeros" (if there are any—and by "trailing zeros", this means any number consecutive zeros/and starting with "the consecutive zeros" only —whether forward (i.e., "zeros following"; or backward (i.e. "zeros preceding").

In our case we have "zeros preceding";  that is a decimal point with zeros PRECEDING an "integer expression"<span>
</span><span> (the "integer" is "175").</span>
______________________________________________________
We then take the "integer expression" (whatever it may be:  12, 5, 30000001 ; or could be a negative value,  etc.) ;  

→  In our case, the "integer expression" is:  "175" ;

and take the first digit (if the expression is negative, we take the negative value of that digit;  if there is only ONE digit (positive or negative), then that is the digit we take ;

And write a decimal point after that first digit (unless in some cases, there is only one digit);  and follow with the rest of the consecutive digits of that 'integer expression' ;

→ In our case:  "175" ; becomes:  " 1.75" .
__________________________________________________
Then we write:  "  * 10^ "
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   {that is "[times]"; or "multiplied by" :    [10 raised exponentially to the power of  <u>     </u> ]._____________________________________________________
 And to find that power, we take the "rewritten integer value (i.e. "whole number value that as been rewritten to a single digit with a decimal point"); and count the [number of "trailing zeros";  if there are any; PLUS the number of decimal places one goes] ; and that number is the value to which "10" is raised.
{If there are none, we write:  " * 10⁰ " ;    since "any value, raised to the "zero power", equals "1" ; so " * 10⁰ " ; is like writing:  " * 1 " .

If there are "trailing zeros" AND/OR or  any number of decimal places,  to the "right" of this expression; the combined number of spaces to the right is: 
  { the numeric value (i.e. positive number) of the power to which "10" is raised }.

Likewise, if there are "trailing zeros" AND/OR or any number of decimal places, to the "LEFT" of this expression; the combined number of spaces to the LEFT is the value of the power which "10" is raised to; is that number—which is a negative value.

In our case:  we have:  0.000000000175 * 10^(-10) .

Note:  The original notation was:

             →  " 0.000000000175 m "

{that is:  "175" [with 9 (nine) zeros to the left].}.

We rewrite the "175" ("integer expression") as:

"1.75" .
____________________________________________________
So we have:
         →   " 0.000000000175 m " ;

Think of this value as:

        " 0. 0000000001{pseudo-decimal point}75   m ".

And count the number of decimal spaces "backward" from the
      "pseudo-decimal point" to the actual decimal; and you will see that there are "10" spaces (to the left).   
______________________________________________________
Also note:  We started with "9 (nine)" preceding "zeros" before the "1" ;  now we are considering the "1" as an "additional digit" ;
             →  "9 + 1 = 10" .
______________________________________________________
Since the decimals (and zeros) come BEFORE (precede) the "175" ; that is, to the "left" of the "175" ; the exponent to which the "10" is raised is:
 "NEGATIVE TEN" { "-10" } .

So we write this value as:  " 1.75 * 10^(-10)  m " .  

{NOTE:  Do not forget the units of measurement; which are "meters" —which can be abbreviateds as:  "m" .} . 
______________________________________________________
The answer is:  " 1.75 * 10^(-10)   m " .
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4 0
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C₄H₉OH + HBr = C₄H₉Br + H2O

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Molar mass of the butanol = 74.12 g/mole

Moles of the alcohol = 1/74.12 = 0.01349 moles

So, moles of bromobutane = 0.01349 moles

Molar mass of C₄H₉Br = 137.018 g/moles

So, theoretical mass of bromobutane is = 0.01349 × 137.0.18

= 1.85 g


6 0
3 years ago
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A silver cube with an edge length of 2.42 cm and a gold cube with an edge length of 2.75 cm are both heated to 85.4 ∘C and place
kakasveta [241]

Answer:

Explanation:

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mass of silver cube = volume x density

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mass of gold cube =  20.8 x 19.3 = 401.44 gm

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87.45 x ( 85.4 - T )

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= 112 x 1 x ( T - 20.5 )

= 112 T - 2296

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7 0
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