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3 years ago

10

Calculate the heat energy released when 24.1g of liquid Mercury at 25 degrees celsius is converted to solid? mercury.; Steve I g

ot that answer wrong. I have to convert grams to moles, find heat released in freezing and cooling and then add the amount of heat released Heat capacity for Hg(l)=28.0J/mol-K enthalpy of fusion: 2.29 kJ/mol

1 answer:

harkovskaia [24]3 years ago

7 0

M<span>olar mass Hg = 200.5920g/mol </span>
<span>Hg = 24.1/200.5920 = 0.120mol </span>

<span>234.32K liquid to 25C (298.15K) liquid </span>
<span>h = (0.120)(28.0)(298.15 - 234.32) = 214.4688 J </span>

<span>234.32K liquid to 234.32K solid </span>
<span>h = (0.120)(2.29) = 0.2748 kj </span>

<span>total heat = 0.2145 + 0.2748 = 0.4893 kJ</span>

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What is the volume, in milliliters, occupied by 30.07 g of an object of density equal to

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$\boxed {\tt 20.317567567568 \ mL}$

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Let's rearrange the formula for $v$ . the volume. Multiply both sides by $v$ , then divide by $d$ .

$d*v=\frac{m}{v}*v$

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The volume can be found by dividing the mass by the density. The mass of the object is 30.07 grams and the density is 1.48 grams per milliliter.

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Divide. Note, when dividing, the grams, or $g$ will cancel out.

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3 years ago

A 3.00-kg block of copper at 23.0°C is dropped into a large vessel of liquid nitrogen at 77.3 K. How many kilograms of nitrogen

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Answer:

1.2584kg of nitrogen boils.

Explanation:

Consider the energy balance for the overall process. There are not heat or work fluxes to the system, so the total energy keeps the same.

For the explanation, the 1 and 2 subscripts will mean initial and final state, and C and N2 superscripts will mean copper and nitrogen respectively; also, liq and vap will mean liquid and vapor phase respectively.

The overall energy balance for the whole system is:

$U_1=U_2$

The state 1 is just composed by two phases, the solid copper and the liquid nitrogen, so: $U_1=U_1^C+U_1^{N_2}$

The state 2 is, by the other hand, composed by three phases, solid copper, liquid nitrogen and vapor nitrogen, so:

$U_2=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

So, the overall energy balance is:

$U_1^C+U_1^{N_2}=U_2^C+U_{2,liq}^{N_2}+U_{2,vap}^{N_2}$

Reorganizing,

$U_1^C-U_2^C=U_{2,liq}^{N_2}+U_{2,vap}^{N_2}-U_1^{N_2}$

The left part of the equation can be written in terms of the copper Cp because for solids and liquids Cp≅Cv. The right part of the equation is written in terms of masses and specific internal energy:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_1^{N_2}u_1^{N_2}$

Take in mind that, for the mass balance for nitrogen, $m_1^{N_2}=m_{2,liq}^{N_2}+m_{2,vap}^{N_2}$ ,

So, let's replace $m_1^{N_2}$ in the energy balance:

$m_C*Cp*(T_1^C-T_2^C)=m_{2,liq}^{N_2}u_{2,liq}^{N_2}+m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,liq}^{N_2}u_1^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

So, as you can see, the term $m_{2,liq}^{N_2}u_{2,liq}^{N_2}$ disappear because $u_{2,liq}^{N_2}=u_{1,liq}^{N_2}$ (The specific energy in the liquid is the same because the temperature does not change).

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}u_{2,vap}^{N_2}-m_{2,vap}^{N_2}u_1^{N_2}$

$m_C*Cp*(T_1^C-T_2^C)=m_{2,vap}^{N_2}(u_{2,vap}^{N_2}-u_1^{N_2})$

The difference $(u_{2,vap}^{N_2}-u_1^{N_2})$ is the latent heat of vaporization because is the specific energy difference between the vapor and the liquid phases, so:

$m_{2,vap}^{N_2}=\frac{m_C*Cp*(T_1^C-T_2^C)}{(u_{2,vap}^{N_2}-u_1^{N_2})}$

$m_{2,vap}^{N_2}=\frac{3kg*0.092\frac{cal}{gC} *(296.15K-77.3K)}{48.0\frac{cal}{g}}\\m_{2,vap}^{N_2}=1.2584kg$

3 0

3 years ago