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3 years ago

A person breathes 2.6 L of air at -11 C into her lungs, where it is warmed to 37 C. What is its new Volume?

Chemistry

1 answer:

aalyn [17]3 years ago

5 0

Answer:

3.076 L.

Explanation:

The pressure is constant as it is the atmospheric pressure.

According to Charles’ law; at constant pressure, the volume of a given quantity of a gas varies directly with its temperature.

∴ V₁T₂ = V₂T₁.

V₁ = 2.6 L & T₁ = -11 °C + 273 = 262 K.

V₂ = ??? L & T₂ = 37 °C + 273 = 310 K.

∴ V₂ = (V₁T₂) / T₁ = (2.6 L)(310 K) / (262 K) = 3.076 L.

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2 years ago

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2 years ago

Within the nuclear us of individual Adams of the same element, the proton number

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the same

Explanation:

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3 years ago

Part a use these data to calculate the heat of hydrogenation of buta-1,3-diene to butane. c4h6(g)+2h2(g)→c4h10(g)

Reptile [31]

Answer: The heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

Explanation:

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H_{rxn}=\sum [n\times \Delta H_{(product)}]-\sum [n\times \Delta H_{(reactant)}]$

For the given chemical reaction:

$C_4H_6(g)+2H_2(g)\rightarrow C_4H_{10}(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H_{rxn}=[(1\times \Delta H_{(C_4H_{10})})]-[(1\times \Delta H_{(C_4H_6)})+(2\times \Delta H_{(H_2)})]$

We are given:

$\Delta H_{(C_4H_{10})}=-2877.6kJ/mol\\\Delta H_{(C_4H_6)}=-2540.2kJ/mol\\\Delta H_{(H_2)}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H_{rxn}=[(1\times (-2877.6))]-[(1\times (-2540.2))+(2\times (-285.8))]\\\\\Delta H_{rxn}=234.2J$

Hence, the heat of hydrogenation of the reaction is coming out to be 234.2 kJ.

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2 years ago

Why do elements that make positive ions occur on the left side of the periodic table while those that make negative ions occur o

Lynna [10]

Answer:

Elements in the same group have the same number of valence electrons.

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3 years ago