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Y_Kistochka [10]

3 years ago

6

Which enthalpy change (î"h) is correct for a substance changing from the liquid phase to the solid phase?

1 answer:

Mnenie [13.5K]3 years ago

7 0

I believe this should be the enthalpy of fusion. Enthalpy change is the amount of heat evolved or absorbed in a reaction. Enthalpy of fusion also called the heat of fusion resulting from providing heat energy or release of energy to a specific amount of a substance to change from solid state to liquid, without change in temperature.

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To test the effectiveness of a gunpowder mixture, 1 gram was exploded under controlled STP conditions, and the reaction chamber

Phoenix [80]

Answer:

The volume of the gas at given temperature is $1300 cm^3$

Explanation:

$P_1$ = initial pressure of gas = 1 atm

$T_1$ = initial temperature of gas = $0^oC=273.15 K$

$V_1$ = initial volume of gas = $310 cm^3 = 310 mL = 0.310 L$

( $1 cm^3= 1mL$ , 1 mL = 0.001 L)

$P_1V_1=nRT_1$ ..[1]

$P_2$ = final pressure of gas = 2.1 atm

$T_2$ = final temperature of gas = $2,200^oC=273+2200 K=2473.15 K$

$V_2$ = final volume of gas = ?

$P_2V_2=nRT_2$ ..[2]

By dividing [1] and [2] we get combined gas equation :,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$V_2=\frac{P_1V_1\times T_2}{T_1\times P_2}$

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

$V_2=\frac{1 atm\times 0.310 L\times 2473.15 K}{273.15 K\times 2.1 atm}$

$V_2=1.3 L=1300 mL= 1300 cm^3$

The volume of the gas at given temperature is $1300 cm^3$

3 0

3 years ago

Read 2 more answers

4.5 centimeters= km<br> please help me girl i’m not trying to fail science

finlep [7]

Answer:

4.5cm = 45/1000000 km

or 4.5cm= 4.5×10^-5 km.

4 0

3 years ago

Read 2 more answers

Sadiq repeated the experiment by adding sulphuric acid to magnesium carbonate. Write the word equation for this reaction below:

anygoal [31]

Answer:

Sulphuric acid + Magnesium Carbonate → Magnesium sulphate + carbon dioxide + water

Explanation:

Sulphuric acid + Magnesium Carbonate

The products are;

Magnesium sulphate, carbon dioxide and water

MgCO3 (s) + H2SO4 (aq) → MgSO4 (aq) + CO2 (g) + H2O (l)

Word Equation;

Sulphuric acid + Magnesium Carbonate → Magnesium sulphate + carbon dioxide + water

7 0

3 years ago

What is the thermal energy needed to completely melt 9.60 mol of ice at 0.0 C

inysia [295]

The thermal energy needed to completely melt 9.60 mole of ice at 0.0 C is 57.8 Kj

Explanation
ice melt to form water

The molar heat of fusion for water is 6.02 Kj/mol
Thermal energy = moles x molar heat of fussion for water

=9.6 mol x6.02 kj/mol =57.8 Kj

4 0

3 years ago

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1. The solubility of lead(II) chloride at some high temperature is 3.1 x 10-2 M. Find the Ksp of PbCl2 at this temperature.

solniwko [45]

Answer:

1) The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2) The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3)The given statement is false.

Explanation:

1)

Solubility of lead chloride = $S=3.1\times 10^-2M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

S 2S

The solubility product of the lead(II) chloride = $K_{sp}$

$K_{sp}=[Pb^{2+}][Cl^-]^2$

$K_{sp}=S\times (2S)^2=4S^3=4\times (3.1\times 10^{-2})^3=1.2\times 10^{-4}$

The solubility product of the lead(II) chloride is $1.2\times 10^{-4}$ .

2)

Concentration of aluminium nitrate = 0.000010 M

Concentration of aluminum ion = $1\timed 0.000010 M=0.000010 M$

Solubility of aluminium hydroxide in aluminum nitrate solution = $S$

$Al(OH)_3(aq)\rightleftharpoons Al^{3+}(aq)+3OH^-(aq)$

S 3S

The solubility product of the aluminium nitrate = $K_{sp}=1.0\times 10^{-33}$

$K_{sp}=[Al^{3+}][OH^-]^3$

$1.0\times 10^{-33}=(0.000010+S)\times (3S)^3$

$S=1.6\times 10^{-10} M$

The solubility of the aluminium hydroxide is $1.6\times 10^{-10} M$ .

3.

$Molarity=\frac{Moles}{Volume (L)}$

Mass of NaCl= 3.5 mg = 0.0035 g

1 mg = 0.001 g

Moles of NaCl = $\frac{0.0035 g}{58.5 g/mol}=6.0\times 10^{-5} mol$

Volume of the solution = 0.250 L

$[NaCl]=\frac{6.0\times 10^{-5} mol}{0.250 L}=0.00024 M$

1 mole of NaCl gives 1 mole of sodium ion and 1 mole of chloride ions.

$[Cl^-]=[NaCl]=0.00024 M$

Moles of lead (II) nitrate = n

Volume of the solution = 0.250 L

Molarity lead(II) nitrate = 0.12 M

$n=0.12 M]\times 0.250 L=0.030 mol$

1 mole of lead nitrate gives 1 mole of lead (II) ion and 2 moles of nitrate ions.

$[Pb^{2+}]=[Pb(NO_2)_3]=0.030 M$

$PbCl_2(aq)\rightleftharpoons Pb^{2+}(aq)+2Cl^-(aq)$

Solubility of lead(II) chloride = $K_{sp}=1.2\times 10^{-4}$

Ionic product of the lead chloride in solution :

$Q_i=[Pb^{2+}][Cl^-]^2=0.030 M\times (0.00024 M)^2=1.7\times 10^{-9}$

$Q_i$ ( no precipitation)

The given statement is false.

3 0

3 years ago