Chemistry, 22.01.2020 20:31 justin5163

A75.0 g piece of gold at 650. k is dropped into 180. g of h2o(l) at 310. k in an insulated container at 1 bar pres- sure. calculate the temperature of the system once equilib- rium has been reached. assume that cp, m for au and h2o is constant at their values for 298 k throughout the temperature range of interest.

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Answer from: electrofy456

The temperature of the system after reaching equilibrium is 314.21 K.

Explanation:

Heat lost by gold will be equal to heat gained by the water

-Q_1=Q_2

Mass of iron = m_1=75.0 g

Specific heat capacity of gold= c_1=0.126 J/gK

Initial temperature of the gold= T_1=650.0 K

Final temperature of gold = T_2 =T

$Q_1=m_1c_1\times (T-T_1)$

Mass of water= m_2=180.0 g

Specific heat capacity of water= c_2=4.184 J/gK

Initial temperature of the water = T_3=310 K

Final temperature of water = T_2 =T

$Q_2=m_2c_2\times (T-T_3)$

-Q_1=Q_2

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

$-(75.0 g\times 0.126 J/gK\times (T-650.0K))=180.0 g\times 4.184 J/gK\times (T-310.0K)$

we get, T = 314.21 K

The temperature of the system after reaching equilibrium is 314.21 K.

Answer from: Quest

the correct option is c.

isotopes refers to different forms of the same element with the same atomic number but different mass number as a result of difference in their neutron numbers. isotopes have different relative atomic mass but share the same chemical properties.

Answer from: Quest

answer : (a) volume of the gas = 8.67 × $10^{-3}$ l

(b) second virial coefficient b = -1.21226 l/mole

solution : given, t = 300k

p = 20atm

z = 0.86

n = 8.2 mmoles = 8.2 × $10^{-3}$ moles

r = 0.082 l atm/k/mol

find, (a) volume of gas and (b) second virial coefficient b at 300k.

formula used : $z=\frac{pv}{nrt}$

and $z=1+\frac{b}{v}$

(a) $z=\frac{pv}{nrt}$

$v=\frac{znrt}{p}$

= $\frac{0.86\times 8.2\times[tex]10^{-3}$ \times 0.082\times 300}{20}[/tex]

= 8.67 × $10^{-3}$ l

(b) $z=1+\frac{b}{v}$

$b=(z-1)v$

= (0.00867-1) × 8.659

= - 8.5839 l/mole

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A75.0 g piece of gold at 650. k is dropped into 180. g of h2o(l) at 310. k in an insulated container...

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