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2 years ago

Compare the amount of thermal energy needed to melt a solid and the amount of thermal energy needed to freeze the same liquid

2 answers:

6 0

To melt a solid, energy must be input into the solid to give the molecules high energy to disrupt the lattice bonds and change phase. To return the liquid to solid, the same amount of energy used in melting is required to be removed from the liquid, ideally. This is assuming there is no loss of energy in the process.

lbvjy [14]2 years ago

3 0

Answer:

They have the same magnitude but differ on the sign.

Explanation:

Hello,

Melting enthalpy is related with the necessary energy a solid sample needs to be converted to liquid; in general, such energy is positive as the initial solid needs to gain energy to become liquid as liquid molecules are arranged farther than solid molecules. On the contrary, freezing enthalpy is related with the energy a liquid sample needs to be converted to solid, in such a way, it is negative since that energy must be withdrawn to promote the liquid molecules to get closer and consequently turn solid. Summing this up, they have the same magnitude but differ on the sign.

Best regards.

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2 years ago

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Suppose a solution is prepared by dissolving 15.0 g NaOH in 0.150 L of 0.250 M nitric acid. What is the final concentration of O

Kipish [7]

Answer:

2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

Explanation:

Moles of NaOH = $\frac{15.0 g}{40 g/mol}=0.375 mol$

Molarity of the nitric acid solution = 0.250 M

Volume of the nitric solution = 0.150 L

Moles of nitric acid = n

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

$n=0.250 M\times 0.150 L=0.0375 mol$

$NaOH+HNO_3\rightarrow NaNO_3+H_2O$

According to reaction, 1 mole of nitric acid recats with 1 mole of NaOH, then 0.0375 moles of nitric acid will react with :

$\frac{1}{1}\times 0.0375 mol$ of NaOH

Moles of NaOH left unreacted in the solution =

= 0.375 mol - 0.0375 mol = 0.3375 mol

$NaOH(aq)\rightarrow Na^+(aq)+OH^-(aq)$

1 mole of sodium hydroxide gives 1 mol of sodium ions and 1 mole of hydroxide ions.

Then 0.3375 moles of NaOH will give :

$1\times 0.3375 moles=0.3375 mol$ of hydroxide ion

The molarity of hydroxide ion in solution ;

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2.25 M is the final concentration of hydroxide ions ions in the solution after the reaction has gone to completion.

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2 years ago

if a liquid is heated and the temperature at which it boils is measured, the ? property is beingm easured.

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You are measuring an Intensive property

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2 years ago

A chemist titrates 160.0mL of a 0.3403M aniline C6H5NH2 solution with 0.0501M HNO3 solution at 25°C . Calculate the pH at equiva

maxonik [38]

Answer : The pH of the solution is, 5.24

Explanation :

First we have to calculate the volume of $HNO_3$

Formula used :

$M_1V_1=M_2V_2$

where,

$M_1\text{ and }V_1$ are the initial molarity and volume of $C_6H_5NH_2$ .

$M_2\text{ and }V_2$ are the final molarity and volume of $HNO_3$ .

We are given:

$M_1=0.3403M\\V_1=160.0mL\\M_2=0.0501M\\V_2=?$

Putting values in above equation, we get:

$0.3403M\times 160.0mL=0.0501M\times V_2\\\\V_2=1086.79mL$

Now we have to calculate the total volume of solution.

Total volume of solution = Volume of $C_6H_5NH_2$ + Volume of $HNO_3$

Total volume of solution = 160.0 mL + 1086.79 mL

Total volume of solution = 1246.79 mL

Now we have to calculate the Concentration of salt.

$\text{Concentration of salt}=\frac{0.3403M}{1246.79mL}\times 160.0mL=0.0437M$

Now we have to calculate the pH of the solution.

At equivalence point,

$pOH=\frac{1}{2}[pK_w+pK_b+\log C]$

$pOH=\frac{1}{2}[14+4.87+\log (0.0437)]$

$pOH=8.76$

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-8.76\\\\pH=5.24$

Thus, the pH of the solution is, 5.24

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2 years ago

If 25.8 grams of BaO dissolve in enough water to make a 212-gram solution, what is the percent by mass of the solution?

Igoryamba

The correct answer is 12.2% BaO.

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2 years ago