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3 years ago

Thermal energy chemical change lab report

Chemistry

1 answer:

natulia [17]3 years ago

5 0

Answer:

This is required answer.

Explanation:

Given that,

Thermal energy chemical change lab report

We know that,

Thermal energy :

When we rise a temperature of any substance then the molecules of substance moves fastly then the heat produced. It is called thermal energy.

Chemical change in the thermal energy :

When we heat a substance then the thermal energy increases and the molecules of substance moves fastly. That means the chemical bonds are break in reactants due to energy.

In products, The new chemical bonds are create due to energy.

For example ; The coal is burned then produced heats water and change it into steam.

Hence, This is required answer.

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Calculate the percent by mass of a solution of Ca(NO3)2 that has 22.63 g dissolved in 896.92 g of water.

never [62]

Hey there!

To calculate the percent by mass of the Ca(NO₃)₂ we need to find the total mass first by adding.

896.92 + 22.63 = 919.55

In total, the solution is 919.55 grams.

To find the percent of Ca(NO₃)₂ in the solution, divide the mass of Ca(NO₃)₂ by the total mass and multiply by 100.

22.63 ÷ 919.55 = 0.0246

0.0246 x 100 = 2.46

Ca(NO₃)₂ makes up 2.46% of the solution.

Hope this helps!

6 0

3 years ago

Which solution has a molality of 0.25m nacl?

Solnce55 [7]

Molality can be expressed by moles of solute over kilograms of solvent. The question asks the molality of 0.25m NaCl. 0.25m NaCl is equal to 0.25 moles of NaCl over 1 kg of water.

5 0

3 years ago

Read 2 more answers

How many moles of silver are there in 6.9 x 10^28 silver atoms?

Juli2301 [7.4K]

Answer:

1.2 x 10⁵ moles Ag (2 sig. figs.)

Explanation:

1 mole any substance (elements or compounds) => 6.023 x 10²³ particles of specified substance

∴ 6.9 x 10²⁸ atoms Ag = 6.9 x 10²⁸ Ag atoms / 6.023 x 10²³ Ag atoms/mole Ag

= 1.145608501 x 10⁵ moles Ag (calculator answer)

= 1.2 x 10⁵ moles Ag (2 sig. figs.)

6 0

2 years ago

Read 2 more answers

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.

6 0

3 years ago

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What is the chemical structure of coal?<br>What is the molecular weight of coal?

nydimaria [60]

1) As for its chemical composition, coal is a mixture of high-molecular-weight polycyclic aromatic compounds, such as benzene C6H6, toluene C6H5CH3, xylene C6H4(CH3)2, naphthalene C10H8, anthracene C14H10, pyrene C16H10 and their derivatives with high mass fraction of carbon, as well as of water and volatile substances.

2) The coal asphaltenes have a relatively narrow MWD (full width ≈ 150 amu) with an average molecular weight of ≈340 amu. The petroleum asphaltenes display a broader MWD (full width ≈ 300 amu) and are heavier on average (≈680 amu).

~ I hope this helped, and I would appreciate Brainliest. ♡ ~ ( I request this to all the lengthy answers I give ! )

4 0

3 years ago