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lora16 [44]
2 years ago
8

How many joules are required to heat 2 l of water in a pot from 20 ◦c to the boiling point of water? the specific heat of water

is 4.18 j g · ◦ c and the density of water is 1 g/ml?
Chemistry
1 answer:
miss Akunina [59]2 years ago
4 0
  The  joules  required  to   heat 2L  of  water  in  a  pot  from  20  c  to  the  boiling  point  of  water  is  calculated     using the following  formula

  Q=    MC  delta  T
M  = mass  =   density  x   volume(  2  x 1000= 2000ml)
M  =     1g/ml  x2000  ml  =  2000g
C  =  specific  heat   capacity =  4.18  g/c
delta  T  =  change  in  temperature  =  100 c (  boiling  point  of  water)   -  20 c =  80 c 
Q  is therefore  =   2000 g x 4.18 j/g c  x  80c  =  668800j
 


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4.6*10^{-14} M

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Constructing an ICE table;we have:

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Initial (M)             0.020          0.40                        0

Change (M)         -  x                - 4 x                       x

Equilibrium (M)  0.020 -x        0.40 - 4 x              x

Given that: K_f =1.7*10^{13}

K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}

1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}

Since x is so small; 0.40 -4x = 0.40

Then:

1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}

1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}

1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x

8.704*10^9-4.352*10^{11}x =x

8.704*10^9 = 4.352*10^{11}x

x = \frac{8.704*10^9}{4.352*10^{11}}

x = 0.0199999999999540

Cu^{2+}= 0.020 - 0.019999999999954

Cu^{2+} = 4.6*10^{-14} M

8 0
2 years ago
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