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4 years ago

When 6.889 grams of a hydrocarbon, CxHy, were burned in a combustion analysis apparatus, 20.17 grams of CO2 and 12.39 grams of H

2O were produced. what is the empirical and molecular formula

Chemistry

1 answer:

Ket [755]4 years ago

6 0

$x^{2} x^{2} \sqrt{x} \left \{ {{y=2} \atop {x=2}} \right. \lim_{n \to \infty} a_n \leq \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \int\limits^a_b {x} \, dx \sqrt{x} x_{123} \leq \leq$

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I would say the correct answer would be Elements.

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3 years ago

State the numbers of protons, neutrons & electrons in an atom of Carbon-14

SashulF [63]

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Neutral carbon-14 contains six protons, eight neutrons, and six electrons; its mass number is 14 (six protons plus eight neutrons). These two alternate forms of carbon are isotopes.

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3 years ago

Top fuel dragsters and funny cars burn nitro-methane as fuel according to the following balanced combustion equation: 2CH3NO2(l)

mart [117]

Answer : The standard enthalpy of formation for nitro-methane is, -467.4 kJ

Explanation :

Enthalpy change : It 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 $\Delta H^o$

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

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$2CH_3NO_2(l)+\frac{3}{2}O_2(g)\rightleftharpoons 2CO_2(s)+3H_2O(g)+N_2(g)$

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

$\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})+(n_{(N_2)}\times \Delta H^o_f_{(N_2)})]-[(n_{(CH_3NO_2)}\times \Delta H^o_f_{(CH_3NO_2)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_{rxn}=-709.2kJ/mol$

$\Delta H^o_f_{(CH_3NO_2(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393kJ/mol\\\Delta H^o_f_{(H_2O(g))}=-286kJ/mol\\\Delta H^o_f_{(N_2(g))}=0kJ/mol$

Putting values in above equation, we get:

$-709.2=[(2\times -393)+(3\times -286)+(1\times 0)]-[(2\times \Delta H^o_f_{(CH_3NO_2)})+(\frac{3}{2}\times 0)]$

$\Delta H^o_f_{(CH_3NO_2)}=-467.4kJ$

Thus, the standard enthalpy of formation for nitro-methane is, -467.4 kJ

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3 years ago

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4 years ago

Choose the word that best completes this sentence. Welding in confined spaces or on stainless steel generates _____________ a ve

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3 years ago

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