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V125BC [204]
3 years ago
8

The specific heat of silver is 0.234 j/(g·c). calculate the amount of energy that is needed to raise the temperature of 175 g of

silver from 22.5°c to 40.0°c.
Chemistry
1 answer:
Elodia [21]3 years ago
3 0
Heat energy is supplied to materials and can cause an increase in temperature of the material. the formula is as follows
H = mcΔt
where H - heat energy 
m - mass of material 
c - specific heat 
Δt - change in temperature  - 40.0 °C - 22.5 °C = 17.5 °C
substituting the values 
H = 175 g x 0.234 Jg⁻¹°C⁻¹ x 17.5 °C
H = 716.6 J 
716.6 J is required 
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0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati
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[AgNO3] = 0.20 M

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Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

BaCrO_4 (s)\leftrightharpoons  Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ba^{2+}][CrO_{4}^{2-}]

1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]

[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}

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Ag_{2}CrO_4(s) \leftrightharpoons  2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)

Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]

1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})

[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}

[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

                       

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