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Ethanol (c2h5oh) melts at -114Â°c. the enthalpy of fusion is 5.02 kj/mol. the specific heats of solid and liquid ethanol are 0.9

Nimfa-mama [501]

Answer: 4.18925 kJ heat is needed to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

Explanation:

Temperature of Solid $C_2H_5OH=-135^oC=138 K(0^oC=273K)$

Melting temperature of Solid $C_2H_5OH=114^oC=159 K$

Temperature of liquid $C_2H_5OH=-50^oC=223K$

Specific heats of solid ethanol = 0.97 J/gK

Specific heats of liquid ethanol = 2.3 J/gK

Heat required to melt the the 25 g solid $C_2H_5OH$ at 159 K

$\Delta T_1$ = 159 K - 138 K = 21 K

$Q_1=mc\Delta T= 25\times 0.97J/gK\times 21 K=509.25 J$

Heat required to melt and raise the temperature of $C_2H_5OH$ upto 223 K

$\Delta T_2$ = 223 K - 159 K = 64 K

$Q_2=mc\Delta T= 25\times 2.3J/gK\times 64 K=3680 J$

Total heat to convert solid ethanol to liquid ethanol at given temperature :

$Q_1+Q_2=509.25 J+3680 J=4189.25 J=4.18925 kJ$ (1kJ=1000J)

Hence, 4.18925 kJ of heat will be required to convert 25.0 g of solid ethanol at -135 °C to liquid ethanol at -50°C.

6 0

2 years ago

The N2O4−NO2 reversible reaction is found to have the following equilibrium partial pressures at 100∘C. Calculate Kp for the rea

timofeeve [1]

Answer:

$K_{p}$ for the reaction is 18.05

Explanation:

Equilibrium constant in terms of partial pressure ( $K_{p}$ ) for this reaction can be written as-

$K_{p}=\frac{P_{NO_{2}}^{2}}{P_{N_{2}O_{4}}}$

where $P_{NO_{2}}$ and $P_{N_{2}O_{4}}$ are equilibrium partial pressure of $NO_{2}$ and $N_{2}O_{4}$ respectively

Hence $K_{p}=\frac{(0.095)^{2}}{(0.0005)}$ = 18.05

So, $K_{p}$ for the reaction is 18.05

3 0

3 years ago

Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. be sure your answer has the correct number

Vera_Pavlovna [14]

<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. 1mmol = 10^-3 mol Therefore 4.10*10^-5mmol = 4.10*10^-8mol molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below) But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g Mass is = 9.75*10^-7 grams 1Âµg = 10^-6 g You then have 9.75*10^-7/10^-6 = 0.975Âµg ZnC2O4 (*see below) at this point you could have said: 1Âµg = 10^-6 g therefore you have a solution of 6.29Âµg per litre, 155ml = 6.29*155/1000 = 0.975Âµg ZnC2O4</span>

3 0

2 years ago

the reaction of aluminum with chlorine gas is shown 2Al + 3Cl2 -> 2AlCl3 based on this equation how many molecules of chlorin

Vanyuwa [196]

45 molecules of chlorine gas (Cl₂) are needed to react with 30 atoms of aluminum (Al)

The balanced equation for the reaction is given below:

2Al + 3Cl₂ —> 2AlCl₃

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

With the above information, we can determine the number of molecules of Cl₂ needed to react with 30 atoms of Al. This can be obtained as follow:

From the balanced equation above,

2 atoms of Al required 3 molecules of Cl₂.

Therefore,

30 atoms of Al will require = $\frac{30 * 3}{2}\\$ = 45 molecules of Cl₂.