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3 years ago

7

A sample of polystyrene, which has a specific heat capacity of 1.880 J.g .Â°C , is put into a calorimeter (see sketch at right)

that contains 300.0 g of water. The polystyrene sample starts off at 94.9 Â°C and the temperature of the water starts off at 22.0 When the temperature of the water stops changing it's 27.7 Â°C. The pressure remains constant at 1 atm. Calculate the mass of the polystyrene sample.

1 answer:

NISA [10]3 years ago

3 0

Answer:

56.6g

Explanation:

Given that :

Mass of water, m1= 300.0 g

Temperature of water, T1= 22°C

Specific heat capacity of water, C= 4.184 J/g°C

Mass of polystyrene, m2=?

Temperature of polystyrene = 94.9

Specific heat capacity of polystyrene, c2= 1.88 J/g°C

Final temperature = 27.7 oC

Heat lost by polystyrene = Heat gained by water

mc(dT) = mc(dT)

m2 * 1.88 * (94.9 - 27.7) = 300 * 4.184 * (27.7 - 22)

m2 * 1.88*67.2 = 300*4.184*5.7

126.336 * m2 = 7154.64

m2 = 7154.64 / 126.336

m2 = 56.631838

Hence, mass of polystyrene = 56.6g

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1.20 x 10^22 molecules NaOH to gram

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Answer:

$\boxed {\boxed {\sf 0.797 \ g \ NaOH}}$

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<u>1. Convert Molecules to Moles</u>

First, we must convert molecules to moles using Avogadro's Number: 6.022*10²³. This tells us the number of particles in 1 mole of a substance. In this case, the particles are molecules of sodium hydroxide.

$\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Multiply by the given number of molecules.

$1.20*10^{22} \ molecules \ NaOH *\frac {6.022*10^{23} \ molecules \ NaOH} {1 \ mol \ NaOH}}$

Flip the fraction so the molecules cancel out.

$1.20*10^{22} \ molecules \ NaOH *\frac {1 \ mol \ NaOH} {6.022*10^{23} \ molecules \ NaOH}}$

$1.20*10^{22} *\frac {1 \ mol \ NaOH} {6.022*10^{23}}}$

$\frac {1.20*10^{22} \ mol \ NaOH} {6.022*10^{23}}}$

$0.0199269345732 \ mol \ NaOH$

<u>2. Convert Moles to Grams</u>

Next, we convert moles to grams using the molar mass.

We must calculate the molar mass using the values on the Periodic Table. Look up each individual element.

Na: 22.9897693 g/mol
O: 15.999 g/mol
H: 1.008 g/mol

Since the formula has no subscripts, we can simply add the molar masses.

NaOH: 22.9897693+15.999+1.008=39.9967693 g/mol

Use this as a ratio.

$\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

Multiply by the number of moles we calculated.

$0.0199269345732 \ mol \ NaOH*\frac {39.9967693 \ g \ NaOH }{1 \ mol \ NaOH}$

The moles of sodium hydroxide cancel.

$0.0199269345732 *\frac {39.9967693 \ g \ NaOH }{1}$

$0.0199269345732 *39.9967693 \ g \ NaOH$

$0.79701300498 \ g \ NaOH$

The original measurement of molecules has 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandth place. The 0 tells us to leave the 7 in the hundredth place.

$0.797 \ g \ NaOH$

1.20*10²² molecules of sodium hydroxide is approximately 0.797 grams.

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The equilibrium constant is $K =0.02867$

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$M_2 O_3 _{(s)} ----> 2M_{(s)} + \frac{3}{2} O_2_{(g)}$

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The Gibbs free energy of the reaction is mathematically represented as

$\Delta G^o_{re} = \sum \Delta G^o _p - \sum G^o _r$

$\Delta G^o_{re} = \sum \Delta G^o _1 - \sum( G^o _2 +G^o _3)$

Substituting values

From the balanced equation

$\Delta G^o_{re} =[ (2 * 0) + (\frac{3}{2} * 0 )] - [1 * - 8.80]$

$\Delta G^o_{re} = 8.80 kJ/mol =8800J/mol$

The Gibbs free energy of the reaction can also be represented mathematically as

$\Delta G^o_{re} = -RTln K$

Where R is the gas constant with a value of $R = 8.314 J/mol \cdot K$

T is the temperature with a given value of $T = 298 K$

K is the equilibrium constant

Now equilibrium constant for a reaction that contain gas is usually expressed in term of the partial pressure of the reactant and products that a gaseous in state

The equilibrium constant for this chemical reaction is mathematically represented as

$K_p =[ P_{O_2}]^{\frac{3}{2} }$

Where $[ P_{O_2}]$ is the equilibrium pressure of oxygen

The p subscript shows that we are obtaining the equilibrium constant using the partial pressure of gas in the reaction

Now equilibrium constant the subject on the second equation of the Gibbs free energy of the reaction

$K = e^{- \frac{\Delta G^o_{re}}{RT} }$

Substituting values

$K= e^{\frac{8800}{8.314 * 298} }$

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