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3 years ago

A compound consists of 75% magnesium and 25% oxygen. Find the empirical formula.

Chemistry

1 answer:

laila [671]3 years ago

8 0

<h2>Answer:</h2>

% Composition of elements:

Magnesium = 75%
Oxygen = 25%

Atomic mass of given elements:

Magnesium = 24 g
Oxygen = 16 g

Now, divide % composition by Atomic mass:

$\footnotesize\implies Mg = \dfrac{\% \: Composition}{Atomic \: mass} = \dfrac{75}{24} = \bf 3.125$

$\footnotesize\implies O = \dfrac{\% \: Composition}{Atomic \: mass} = \dfrac{25}{16} = \bf 1.5625$

Simplest Ratio:

$\footnotesize\implies Mg = \dfrac{3.125}{1.5625} = 2$

$\footnotesize\implies O = \dfrac{1.5625}{1.5625} = 1$

Empirical Formula:

$\footnotesize\implies \underline{ \boxed{ \red{ \bf Empirical \: Formula = Mg_2O}}}$

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Calculate the entropy change for the surroundings of the reaction below at 350K: N2(g) + 3H2(g) -> 2NH3(g) Entropy data: NH3

krek1111 [17]

Answer : The entropy change for the surroundings of the reaction is, -198.3 J/K

Explanation :

We have to calculate the entropy change of reaction $(\Delta S^o)$ .

$\Delta S^o=S_{product}-S_{reactant}$

$\Delta S^o=[n_{NH_3}\times \Delta S^0_{(NH_3)}]-[n_{N_2}\times \Delta S^0_{(N_2)}+n_{H_2}\times \Delta S^0_{(H_2)}]$

where,

$\Delta S^o$ = entropy of reaction = ?

n = number of moles

$\Delta S^0{(NH_3)}$ = standard entropy of $NH_3$

$\Delta S^0{(H_2)}$ = standard entropy of $H_2$

$\Delta S^0{(N_2)}$ = standard entropy of $N_2$

Now put all the given values in this expression, we get:

$\Delta S^o=[2mole\times (192.5J/K.mole)]-[1mole\times (191.5J/K.mole)+3mole\times (130.6J/K.mole)]$

$\Delta S^o=-198.3J/K$

Therefore, the entropy change for the surroundings of the reaction is, -198.3 J/K

4 0

3 years ago

How are quantum numbers related to atomic orbitals?

sdas [7]

When considering atomic orbitals the only important information they really wanted to know is the size of the orbit, which was described by using quantum numbers.

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3 years ago

Name two compounds with more exothermic lattice energies than scandium oxide and justify your choice.

N76 [4]

Answer:

1 . $Al_2O_3$

2. $TiO_2$

Explanation:

The more stable the ionic compound, the more is it lattice energy.

The more the charge on the cation and the anion, the greater is the lattice energy.
The less the size of the cation and the anion, the greater is the lattice energy.

Scandium oxide ( $Sc_2O_3$ ) is an oxide in which $Sc^{3+}$ behaves as cation and $O^{2-}$ behaves as anion.

The compounds which has higher lattice energy than scandium oxide are:

1 . $Al_2O_3$

This is because the charge are same on the cation and the anion as in the case of the Scandium oxide but the size of the cation $Al^{3+}$ is smaller than $Sc^{3+}$ . Thus, this corresponds to higher lattice energy.

2. $TiO_2$

This is because the charge on the cation $Ti^{4+}$ is greater than that of $Sc^{3+}$ and also the size of the cation $Ti^{4+}$ is smaller than $Sc^{3+}$ . Thus, this corresponds to higher lattice energy.