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3 years ago

12

____ is a poor conductor of heat A. aluminum B.iron C. ceramic D. zinc

1 answer:

Ganezh [65]3 years ago

8 0

Your answer is ceramic.

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A gas mixture with 4 mol of Ar, x moles of Ne, and y moles

maks197457 [2]

Answer:

a) $\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

b) $x=4$

c) $\Delta G_{max}=-2721.9 J/mol$

Explanation:

Gas mixture:

$n_{Ar}= 4 mol$

$n_{Ne}= x mol$

$n_{Xe}= y mol$

$n_{tot}= n_{Ar} + n_{Ne} + n_{Xe}=3*n_{Ar}$

$n_{Ne} + n_{Xe}=2*n_{Ar}$

$x + y=8 mol$

$y=8 mol- x$

Mol fractions:

$x_{Ar}=\frac{4 mol}{12 mol}=1/3$

$x_{Ne}=\frac{x mol}{12 mol}=x/12$

$x_{Xe}=\frac{8 - x mol}{12 mol}=(8-x)/12$

Expression of $\Delta G_{mixing}$

$\Delta G_{mixing}=R*T*\sum_{i]*x_i*ln (x_i)$

$\Delta G_{mixing}=R*T*[1/3*ln (1/3) +x/12*ln (x/12) +(8-x)/12*ln ((8-x)/12)]$

$\Delta G_{mixing}=\frac{R*T}{12}*[4*ln (1/3) +x*ln (x/12) +(8-x)*ln ((8-x)/12)]$

Expression of $\Delta G_{max}$

$\frac{d \Delta G_{mixing}}{dx}=0$

$\frac{d \Delta G_{mixing}}{dx}=\frac{R*T}{12}*[ln (x/12)+12-ln ((8-x)/12)-12]$

$0=\frac{R*T}{12}*[ln (x/12)-ln ((8-x)/12)$

$0=[ln (x/12)-ln ((8-x)/12)$

$ln (x/12)=ln ((8-x)/12)$

$x=(8-x)$

$x=4$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (4/12) +(8-4)*ln ((8-4)/12)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[4*ln (1/3) +4*ln (1/3) +(4)*ln (1/3)]$

$\Delta G_{max}=\frac{8.314*298}{12}*[12*ln (1/3)]$

$\Delta G_{max}=-2721.9 J/mol$

4 0

3 years ago

A gold ring with a mass of 16g was dropped in the snow and its temperature dropped from 35°C to 0°C. How much heat was released

Luda [366]

Answer:

-72.8 joules

Explanation:

just finished the test

4 0

3 years ago

Which of the following is insoluble in water?

bija089 [108]

Answer:

I think Is A

Explanation:

5 0

3 years ago

NaHCO3 (s) + HC2H3O2 (aq) = NaC2H3O2 (aq) + H2O (I) + CO2 (g)

Misha Larkins [42]

Moles=volume*concentration
=0.1*.83
=.083 Moles of HC2H3O2
Mole ratio between HC2H3O2 and CO2 is 1:1
This means .083 Moles of CO2

Mass =Moles*Rfm of CO2
=.083*(12+16+16)
=3.7grams

8 0

4 years ago

Calculate the second volumes. 51.7 L at 27 C and 90.9 KPa to STP

Arturiano [62]

The second volume : 42.2 L

<h3>Further explanation</h3>

Given

51.7 L at 27 C and 90.9 KPa

Required

The second volume

Solution

STP = P₂=1 atm, T₂=273 K

T₁ = 27 + 273 = 300 K

P₁ = 90.9 kPa = 0,897 atm

Use combine gas law :

P₁V₁/T₁ = P₂V₂/T₂

Input the value :

0.897 x 51.7/300 = 1 x V₂/273

V₂= 42.20 L

6 0

3 years ago