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liubo4ka [24]
3 years ago
8

How much energy is required to heat 40g of water from -7 degrees Celsius to 108 degrees Celsius (Lf= 335,000J/Kg for ice)

Chemistry
1 answer:
solniwko [45]3 years ago
6 0
The   energy  required  to heat  40g  of water  from -7 c  to 108 c is
1541000  joules

     calculation

Q(heat)=  M( mass)  x c(specific heat capacity) xdelta t( change in temperature)

M=  40g=  40/1000= 0.04 Kg
C=  335,000 j/kg/c
delta T   (    108 --7= 115  c)

Q  is therefore   =  0.04 g x  335000 j/kg/c  x 115 c  = 1541,000  joules


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never [62]

Answer: B. 1:2

Explanation: Beryllium and chlorine forms a binary ionic compound. Ionic compound is formed when a metal loses its electrons to a receiving non metal. Beryllium (metal) has two valence electrons while chlorine (nonmetal) has seven valence electrons, and so a beryllium atom has to give out its two valence electrons to attain a duplet stable structure while a chlorine atom will gain one electron to attain its stable octet structure. In the reaction between beryllium and chlorine, two atoms of chlorine have to accept the two electrons from one beryllium atom to attain their stable octet structure.

The formula of the compound formed is BeCl2.

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El gas dióxido de carbono ocupa un volumen de 10, 5 litros a 20°C ¿Cuál será el volumen que ocupará el gas si la temperatura Cel
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2 years ago
Whenever a continuous bell alarm sounds in the Chemistry building:
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Answer:

C

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4 0
3 years ago
The copper(I) ion forms a chloride salt (CuCl) that has Ksp = 1.2 x 10-6. Copper(I) also forms a complex ion with Cl-:Cu+ (aq) +
Mnenie [13.5K]

Answer: (a) The solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b) The solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

Explanation:

(a)  Chemical equation for the given reaction in pure water is as follows.

           CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq)

Initial:                         0            0

Change:                    +x           +x

Equilibm:                   x             x

K_{sp} = 1.2 \times 10^{-6}

And, equilibrium expression is as follows.

          K_{sp} = [Cu^{+}][Cl^{-}]

       1.2 \times 10^{-6} = x \times x

             x = 1.1 \times 10^{-3} M

Hence, the solubility of CuCl in pure water is 1.1 \times 10^{-3} M.

(b)  When NaCl is 0.1 M,

       CuCl(s) \rightarrow Cu^{+}(aq) + Cl^{-}(aq),  K_{sp} = 1.2 \times 10^{-6}

   Cu^{+}(aq) + 2Cl^{-}(aq) \rightleftharpoons CuCl_{2}(aq),  K = 8.7 \times 10^{4}

Net equation: CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

               K' = K_{sp} \times K

                          = 0.1044

So for, CuCl(s) + Cl^{-}(aq) \rightarrow CuCl_{2}(aq)

Initial:                     0.1                 0

Change:                -x                   +x

Equilibm:            0.1 - x                x

Now, the equilibrium expression is as follows.

              K' = \frac{CuCl_{2}}{Cl^{-}}

         0.1044 = \frac{x}{0.1 - x}

              x = 9.5 \times 10^{-3} M

Therefore, the solubility of CuCl in 0.1 M NaCl is 9.5 \times 10^{-3} M.

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Vegetables are the seeds.

Fruits are the seeds.

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