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victus00 [196]
2 years ago
6

A 20.0g sample of metal with a specific heat of 5 J/(g°C) raised it's initial temperature to 40.0 when 500J heat was added. What

was the initial temperature of the metal?
Chemistry
1 answer:
slavikrds [6]2 years ago
3 0

The initial temperature of the metal = 35 °C

<h3>Further explanation</h3>

Heat can be formulated :

Q = m . c . ΔT

Q = heat, J

c = specific heat, J/g C

ΔT = temperature, °C

m = 20 g

c = 5 J/(g°C)

Q = 500 J

T₁ = 40 C

the initial temperature :

\tt \Delta t(T_2-T_1)=\dfrac{Q}{m.c}\\\\40-T_1=\dfrac{500}{20.5}\\\\40-T_1=5\\\\T_1=35^oC

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Explanation:

PbBr_{2} will dissociate into ions as follows.

         PbBr_{2}(s) \rightleftharpoons Pb^{2+}(aq) + 2Br^{-}(aq)

Hence, K_{sp} for this reaction will be as follows.

                   K_{sp} = [Pb^{2+}][Br^{-}]^{2}

We take x as the molar solubility of PbBr_{2} when we dissolve x moles of solution per liter.

Hence, ionic molarities in the saturated solution will be as follows.

               [Pb^{2+}] = [Pb^{2+}]_{o} + x

               [Br^{-}]^{2} = [Br^{-}]_{o} + 2x

So, equilibrium solubility expression will be as follows.

            K_{sp} = ([Pb^{2+}]_{o} + x)([Br^{-}]_{o} + 2x)^{2}

Each sodium bromide molecule is giving one bromide ion to the solution. Therefore, one solution contains [Br^{-}]_{o} = 0.10 and there will be no lead ions. So, [Pb^{2+}]_{o} = 0

So, [Br^{-}]_{o} + 2x will approximately equals to [Br^{-}]_{o}.

Hence, K_{sp} = x[Br^{-}]^{2}_{o}

            4.67 \times 10^{-6} = x \times (0.10)^{2}

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Collect like terms

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Collect like terms

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ʸ ₓZ => ⁰₋₁Z => ⁰₋₁B

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