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abruzzese [7]
3 years ago
15

The standard internal energy change for a reaction can be symbolized as Δ U ∘ rxn or Δ E ∘ rxn . For each reaction equation, cal

culate the energy change of the reaction at 25 ∘ C and 1.00 bar . Sn ( s ) + 2 Cl 2 ( g ) ⟶ SnCl 4 ( l ) Δ H ∘ rxn = − 511.3 kJ/mol
Chemistry
1 answer:
rosijanka [135]3 years ago
6 0

Answer : The internal energy change is, -506.3 kJ/mol

Explanation :

Formula used :

\Delta H=\Delta U+\Delta n_gRT

or,

\Delta U=\Delta H-\Delta n_gRT

where,

\Delta H = change in enthalpy = -511.3kJ/mol=-511300.0J/mol

\Delta U = change in internal energy = ?

\Delta n_g = change in moles

Change in moles = Number of moles of product side - Number of moles of reactant side

According to the reaction:

Change in moles = 0 - 2 = -2 mole

That means, value of \Delta n_gRT = 0

R = gas constant = 8.314 J/mol.K

T = temperature = 25^oC=273+25=298K

Now put all the given values in the above formula, we get

\Delta U=\Delta H-\Delta n_gRT

\Delta U=(-511300.0J/mol)-[-2mol\times 8.314J/mol.K\times 298K

\Delta U=-511300.0J/mol+4955.144J/mol

\Delta U=-506344.856J/mol=-506.3kJ/mol

Therefore, the internal energy change is -506.3 kJ/mol

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A gas at 89C occupies a volume of 0.67 L. At least what temperature will the volume increase to 1.12 L
dybincka [34]

At temperature 332.13 °C

<h3>Further explanation</h3>

Charles's Law  stated that :

When the gas pressure is kept constant, the gas volume is proportional to the temperature  

\tt \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}

T₁ = 89 °C=89+273=362 K

V₁ = 0.67 L

V₂ = 1.12 L

\tt T_2=\dfrac{V_2.T_1}{V_1}\\\\T_2=\dfrac{1.12\times 362}{0.67}=605.13~K=332.13~^oC

7 0
3 years ago
Two significance of melting point
eimsori [14]

The significance of melting points are

  • it gives us a idea about inter molecular force which binds the particles together.
  • at melting point the solid and liquid states exist in equilibrium.

<h3>What is melting point?</h3>

The melting point is the temperature at which a given substance change its physical state from solid to liquid. At this point solids and liquids exist at equilibrium. Melting point of a substance depends on pressure. The melting point of ice increases when pressure decreases. Melting point of a substance decreases by the presence of impurities in it. so you increase or decrease the melting point on any substance by adding more impurities. Ionic bonds, shape and size of molecules are some other factors that affect the melting point.

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5 0
2 years ago
the specific heat of the metallic element rhodium is 0.240 j/g-k at 25°c. if 586 j of heat is added to an 80.0 g block of rhodiu
andrey2020 [161]

The expected final temperature of the block, given that 586 J of heat were added to it is 55.5 °C

<h3>How to determine the final temeprature</h3>

We'll begin by obtaining the change in the temperature of the block. This can be obtained as follow:

  • Specific heat capacity of block (C) = 0.240 J/gºC
  • Heat added (Q) = 586 J
  • Mass of block (M) = 80.0 g
  • Change in temperature (ΔT) =?

Q = MCΔT

Divide both sides by MC

ΔT = Q / MC

ΔT = 586 / (80.0 × 0.240)

ΔT = 586 / 19.2

ΔT = 30.5 °C

Finally, we shall determine the final temperature of the block. This can be obtained as follow:

  • Initial temperature (T₁) = 25 °C
  • Change in temperature (ΔT) = 30.5 °C
  • Final temperature (T₂) = ?

ΔT = T₂ – T₁

30.5 = T₂ – 25

Collect like terms

T₂ = 30.5 + 25

T₂ = 55.5 °C

Thus, from the calculation made above, we can conclude that the final temperature is 55.5 °C

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6 0
1 year ago
A balloon contains 2.0 L of air at 101.5 kPa . You squeeze the balloon to a volume of 0.25 L.
8_murik_8 [283]
<h3>Answer:</h3>

812 kPa

<h3>Explanation:</h3>
  • According to Boyle's law pressure and volume of a fixed mass are inversely proportional at constant absolute temperature.
  • Mathematically, P\alpha \frac{1}{V}

At varying pressure and volume;

P1V1=P2V2

In this case;

Initial volume, V1 = 2.0 L

Initial pressure, P1 = 101.5 kPa

Final volume, V1 = 0.25 L

We are required to determine the new pressure;

P2=\frac{P1V1}{V2}

Replacing the known variables with the values;

P2=\frac{(101.5)(2.0L)}{0.25L}

           = 812 kPa

Thus, the pressure of air inside the balloon after squeezing is 812 kPa

8 0
3 years ago
Wat is significant figures
Sav [38]

<em>Answer:</em>

  •                   0.052301 km have 5 significant figure
  •                   400 cm have 1 significant figure
  •                   50.0 m  have 3 significant figure
  •                  4500.01 ml have 6 significant figure

<em>Explanation:</em>

According to rules of significant figure

0.052301 km have 5 significant figure:

  • Zero to the left of the first non zero digit not significant.
  • Zero between the non zero digits are significant.

<em>400 cm have 1 significant figure:</em>

  • Trailing zeros are not significant in numbers without decimal points.

<em>50.0 m  have 3 significant figure:</em>

  • Trailing zeros are significant in numbers when there is decimal points.

<em>4500.01 ml have 6 significant figure:</em>

  • Zero between the non zero digits are significant.
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4 years ago
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