How does a lithium cation compare to a lithium atom?

Chemistry

1 answer:

ELEN [110]3 years ago

6 0

A lithium cation in terms of atomic/ionic size is smaller than a Lithium atom.

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1 C8H10(l) +21/2O2(g) → 8CO2(g) + 5H2O(g), Hcomb= ? Hf for C8H10(l) = +49.0kJ/mol C8H10(l) Use the balanced combustion reaction

nikklg [1K]

Answer:

$H_{comb}=-4406kJ/mol$

Explanation:

Hello,

In this case, the enthalpy of combustion is understood as the energy released when one mole of fuel, in this case octene, is burned in the presence of oxygen and is computed with the enthalpies of formation of the fuel, carbon dioxide and water as shown below (oxygen is circumvented as it is a pure element):

$H_{comb}=8*\Delta _fH_{CO_2}+5\Delta _fH_{H_2O}-\Delta _fH_{C_8H_{10}}$

Thus, since we already know the enthalpy of combustion of the fuel, for carbon and water we have -393.5 and -241.8 kJ/mol respectively, thereby, the enthalpy of combustion turns out:

$H_{comb}=8*(-393.5kJ/mol)+5(-241.8kJ/mol)-49.0kJ/mol\\\\H_{comb}=-4406kJ/mol$

Best regards.

4 0

3 years ago

A small hole in the wing of a space shuttle requires a 15.9 cm2 patch.

Rzqust [24]

The patch area in square kilometers is 1.59*10⁻⁹ km₂

Why?

This is an unit conversion problem. We have to convert from cm² to km². We can do that by knowing that there are 100 cm in 1 m, and 1000 m in 1 km, so 100000 cm=1km. Knowing that we can apply the following conversion factor:

$15.9 cm^{2} * (\frac{1 km}{100000 cm} )^{2}=0.00000000159 km^{2}$

Now to convert this value to scientific notation, we have to move the decimal point to the right until we get a whole number, and the exponent of the number 10 is going to be the number of spaces we moved to the right (negative), so the final answer is:

$0.00000000159 km^{2}=1.59*10^{-9} km^{2}$