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3 years ago

A gas at 200 K occupies a volume of 350 ml. What temperature is needed to increase the volume

Chemistry

1 answer:

Rudik [331]3 years ago

5 0

Answer: Temperature in constant pressure is 286 K

Explanation: If pressure remains constant, then V/T = constant.

V1 = 350 ml and T1 = 200 K and V2 = 500 ml. V1/T1 = V2/T2

and T2 = T1· V2 / V1 = 200 K · 500 ml / 300 ml = 285,7 K

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3 years ago

Describe the changed that happened in the crayon and candle when they cooled after 3 to 5 minutes.

AleksandrR [38]

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3 years ago

How much energy must be removed from a 94.4 g sample of benzene (molar mass= 78.11 g/mol) at 322.0 K to solidify the sample and

Kay [80]

Answer : The energy removed must be, 29.4 kJ

Explanation :

The process involved in this problem are :

$(1):C_6H_6(l)(322K)\rightarrow C_6H_6(l)(279K)\\\\(2):C_6H_6(l)(279K)\rightarrow C_6H_6(s)(279K)\\\\(3):C_6H_6(s)(279K)\rightarrow C_6H_6(s)(205K)$

The expression used will be:

$Q=[m\times c_{p,l}\times (T_{final}-T_{initial})]+[m\times \Delta H_{fusion}]+[m\times c_{p,s}\times (T_{final}-T_{initial})]$

where,

$Q$ = heat released for the reaction = ?

m = mass of benzene = 94.4 g

$c_{p,s}$ = specific heat of solid benzene = $1.51J/g^oC=1.51J/g.K$

$c_{p,l}$ = specific heat of liquid benzene = $1.73J/g^oC=1.73J/g.K$

$\Delta H_{fusion}$ = enthalpy change for fusion = $-9.8kJ/mol=-\frac{9.8\times 1000J/mol}{78g/mol}=-125.6J/g$

Now put all the given values in the above expression, we get:

$Q=[94.4g\times 1.73J/g.K\times (279-322)K]+[94.4g\times -125.6J/g]+[94.4g\times 1.51J/g.K\times (205-279)K]$

$Q=-29427.312J=-29.4kJ$

Negative sign indicates that the heat is removed from the system.

Therefore, the energy removed must be, 29.4 kJ

3 0

3 years ago

How many grams of NaCl (molecular weight = 58.4 g mole-1) would you dissolve in water to make a total volume of 500 mL of soluti

kolezko [41]

Given :

Volume , V = 500 mL .

Molarity , M = 0.5 M .

Molecular mass of NaCl is $58.4\ g/mole$ .

To Find :

How many grams of NaCl is required .

Solution :

Let , NaCl required is x gram .

Molarity is given by :

$M=\dfrac{\text{Number of moles}}{\text{Volume (in liters) }}\\\\M=\dfrac{m}{M\times V}\\\\0.5=\dfrac{x}{58.4\times 0.5}\\\\x=0.5^2\times 58.4\ g\\\\x=14.6\ g$

Hence , this is the required solution.

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2 years ago