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3 years ago

12

Sharping a pencil leaves behind pencil shavings. Is sharpening pencil a physical or chemical change

1 answer:

Deffense [45]3 years ago

6 0

Physical, since you are physically changing its appearance

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Is NaHCO3 a compound? Explain pls!

Slav-nsk [51]

Answer:

Yes it is

Explanation: because it is composed of identical molecules consisting of atoms of two or more chemical elements.

hope it helps <3

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3 years ago

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An object measures 6.2cm x 13.7cm x 26.9cm. Which value is the length of the object?

kotykmax [81]

Answer:

Explanation:

The length is usually the longest side.

That would mean that 26.9 is the length.

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3 years ago

22 Which of the following statements is true for all loving organisms? A. All living organisms are made of one or more cells wit

Klio2033 [76]

The answer is C. No all cells have cell walls. Prokaryotes don’t have a nucleus or chloroplast. Leaving C to be correct

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3 years ago

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A teacher is discussing several scenarios in the topic of gas law s. Which of the following scenarios will the initial pressure

sukhopar [10]

The scenario whereby the initial pressure is equal to the final pressure is : ( A ) A balloon filled with air is heated from 300k to 600k, causing it to expand and double in size.

<h3>Gay Lussac's law </h3>

Gay Lussac's law states that at constant pressure, the volume of gas is directly proportional to the change in temperature. i.e. V ∝ T. From scenario 1 the change in temperature from 300 K to 600 K shows a 100% increase in temperature which corresponds with the expansion and size doubling of the volume of gas.

Hence we can conclude that The scenario whereby the initial pressure is equal to the final pressure is : ( A ) A balloon filled with air is heated from 300k to 600k, causing it to expand and double in size.

Learn more about Gay Lussac's law : brainly.com/question/24691513

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2 years ago

Be sure to answer all parts. The standard enthalpy of formation and the standard entropy of gaseous benzene are 82.93 kJ/mol and

skelet666 [1.2K]

Answer : The values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

Explanation :

The given balanced chemical reaction is,

$C_6H_6(l)\rightarrow C_6H_6(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{C_6H_6(g)}\times \Delta H_f^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta H_f^0_{(C_6H_6(l))}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0_{(C_6H_6(g))}$ = standard enthalpy of formation of gaseous benzene = 82.93 kJ/mol

$\Delta H_f^0_{(C_6H_6(l))}$ = standard enthalpy of formation of liquid benzene = 49.04 kJ/mol

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

$\Delta H^o=[1mole\times (82.93kJ/mol)]-[1mole\times (49.04J/mol)]$

$\Delta H^o=33.89kJ/mol=33890J/mol$

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{C_6H_6(g)}\times \Delta S^0_{(C_6H_6(g))}]-[n_{C_6H_6(l)}\times \Delta S^0_{(C_6H_6(l))}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0_{(C_6H_6(g))}$ = standard entropy of formation of gaseous benzene = 269.2 J/K.mol

$\Delta S^0_{(C_6H_6(l))}$ = standard entropy of formation of liquid benzene = 173.26 J/K.mol

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

$\Delta S^o=[1mole\times (269.2J/K.mol)]-[1mole\times (173.26J/K.mol)]$

$\Delta S^o=95.94J/K.mol$

Now we have to calculate the Gibbs free energy of reaction $(\Delta G^o)$ .

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

At room temperature, the temperature is $25^oC\text{ or }298K$ .

$\Delta G^o=(33890J)-(298K\times 95.94J/K)$

$\Delta G^o=5299.88J/mol=5.299kJ/mol$

Therefore, the values of $\Delta H^o,\Delta S^o\text{ and }\Delta G^o$ are $33.89kJ,95.94J/K\text{ and }5.299kJ/mol$ respectively.

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3 years ago

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