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3 years ago

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9. Calculate the specific heat capacity of titanium if a 43.56 g sample absorbs 0.476 kJ as its temperature changes from 20.5 oC

to 41.2 oC.

1 answer:

Shalnov [3]3 years ago

8 0

Answer:

c = 0.528 J/g.°C

Explanation:

Given data:

Mass of titanium = 43.56 g

Heat absorbed = 0.476 KJ = 476 j

Initial temperature = 20.5°C

Final temperature = 41.2°C

Specific heat capacity = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 41.2°C - 20.5°C

ΔT = 20.7 °C

476 J = 43.56 g × c × 20.7 °C

476 J = 901.692 g.°C × c

c = 476 J / 901.692 g.°C

c = 0.528 J/g.°C

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Explanation:

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Answer:

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Pressure of XO3 = 0.5 atm

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$K_{p} = \frac{[PXO_{3}] ^{2} }{[PXO_{2}] ^{2}[PO_{2}] }$

$K_{p} = \frac{0.5^2}{2.5^2 *0.75} \\K_{p} = 0.0533 = K_{eq}$

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$K_{-1} = \frac{K_{1} }{K_{c} } \\K_{-1} = \frac{7.8 * 10^{-2} }{1.313 }\\K_{-1} = 0.0594 m^{-1} s^{-1}$

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slega [8]

Answer:

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Explanation:

Hello,

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$pH=-log([H^+])$

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(b)

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Nevertheless, for the strong sodium hydroxide, we don't directly compute the pH but the pOH since the concentration of base equals the concentration hydroxyl in the solution:

$[OH^-]=[NaOH]$

$pOH=-log([OH^-])$

$pH=14-pOH$

Thus, we have:

(b)

$pOH=-log(0.1)=1\\pH=14-1=13$

(d)

$pOH=-log(0.05)=1.3\\pH=14-1.3=12.7$

Best regards.

5 0

3 years ago