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Shkiper50 [21]
3 years ago
9

Which of these statements are true? Select all that apply.

Chemistry
2 answers:
Anarel [89]3 years ago
4 0
In order to find the two statements, we must first define what the enthalpy of formation and the enthalpy of reaction mean.

Enthalpy of formation:
The change in enthalpy when one mole of substance is formed from its constituent elemetns at standard state.

Enthalpy of reaction:
The change in enthalpy when a reaction occurs and the reactants and products are in their standard states.

Now, we check the statements. The true ones are:

The Hrxn for C(s) + O₂(g) → CO₂(g) is the same as Hf for CO₂
This is true because the formation of carbon dioxide requires carbon and oxygen in their standard states.

T
he Hf for Br₂<span>(l) is 0 kJ/mol by definition.
Because the bromine is present in its standard state, the enthalpy of formation is 0.

</span><span>The Hrxn for the reaction 1.5H</span>₂<span>(g) + 0.5N</span>₂<span>(g) </span>→ <span>NH</span>₃<span>(g) is the same as the Hf for NH</span>₃<span>(g)
The reactants and products are present in their standard state, and the reaction is the same as the one occurring during the formation of ammonia.
</span>
morpeh [17]3 years ago
3 0

Answer:

1,3,6

Explanation:

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A volume of 105 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If
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Answer:

<h3>25.0 grams is the mass of the steel bar.</h3>

Explanation:

Heat gained by steel bar will be equal to heat lost by the water

Q_1=-Q_2

Mass of steel= m_1

Specific heat capacity of steel = c_1=0.452 J/g^oC

Initial temperature of the steel = T_1=2.00^oC

Final temperature of the steel = T_2=T=21.50^oC

Q_1=m_1c_1\times (T-T_1)

Mass of water= m_2= 105 g

Specific heat capacity of water=c_2=4.18 J/g^oC

Initial temperature of the water = T_3=22.00^oC

Final temperature of water = T_2=T=21.50^oC

Q_2=m_2c_2\times (T-T_3)-Q_1=Q_2(m_1c_1\times (T-T_1))=-(m_2c_2\times (T-T_3))

On substituting all values:

(m_1\times 0.452 J/g^oC\times (21.50^o-2.00^oC))=-(105 g\times 4.18 J/g^oC\times (21.50^o-22.00^o))\\\\m_1*8.7914=241.395\\\\m_1=\frac{219.45}{8.7914} \\\\m_1=24.9\\\\ \approx25 \texttt {grams}

<h3>25.0 grams is the mass of the steel bar.</h3>
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