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3 years ago

A compound has a chemical composition of .62 g carbon, .10 g hydrogen, .28 g oxygen. What is the empirical formula?

1 answer:

4 0

First we divide the masses by molar masses:
0.62 g carbon / (12.01 g/mol carbon) = 0.0516 mol C
0.10 g hydrogen / (1.01 g/mol H) = 0.099 mol H
0.28 g oxygen / (14 g/mol O) = 0.02 mol O
Since O has the smallest amount, we divide the other values by it:
0.0516 mol C / 0.02 mol O = 2.58 ~ 2.5
0.099 mol H / 0.02 mol O = 4.95 ~ 5
Therefore we have a ratio of C2.5, H5, and O. Since an empirical formula must only contain integers, we multiply everything by two to get:
C5H10O2

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Explanation:

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Given that ΔH = −571.6 kJ/mol for the reaction 2 H2(g) + O2(g) → 2 H2O(l), calculate ΔH for these reactions. (a) 2 H2O(l) → 2 H2

pashok25 [27]

Answer : The value of $\Delta H$ for the reaction is +571.6 kJ/mole.

Explanation :

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given chemical reaction is,

$2H_2(g)+O_2(g)\rightarrow 2H_2O(l)$ $\Delta H_1=-571.6kJ/mole$

Now we have to determine the value of $\Delta H$ for the following reaction i.e,

$2H_2O(l)\rightarrow 2H_2(g)+O_2(g)$ $\Delta H_2=?$

According to the Hess’s law, if we reverse the reaction then the sign of $\Delta H$ change.

So, the value $\Delta H_2$ for the reaction will be:

$\Delta H_2=-(-571.6kJ/mole)$

$\Delta H_2=+571.6kJ/mole$

Hence, the value of $\Delta H$ for the reaction is +571.6 kJ/mole.

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In his experiment, Rutherford shot alpha particles towards a very thin sheet of gold foil. He observed the following things:

1- Most of the alpha particles went undeflected, but

2- Some of them were scattered at very large angles

3- A few of them were even reflected back to their original directions

Observations 2) and 3) were incompatible with Thomson model of the atom: in fact, if this model was true, all the alpha particle should have gone undeflected, or scattered at very small angles. Instead, due to observations 2) and 3), it was clear that:

- The positive charge of the atom was all concentred in a tiny nucleus

- Most of the mass of the atom was also concentrated in the nucleus

So, Rutherford experiment lead to a change in the atomic model of the atom, as it was clear that the plum-pudding model was no longer adequate to describe the results of Rutherford's experiment.

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