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3 years ago

What is the empirical formula for a compound whose molecular formula is P4O10

Chemistry

1 answer:

Bas_tet [7]3 years ago

6 0

Answer:

The empirical formula is P2O5 (option B)

Explanation:

An empirical formula does not necessarily represent the actual numbers of atoms present in a molecule of a compound; it represents only the ratio between those numbers.

The actual numbers of atoms of each element that occur in the smallest freely existing unit or molecule of the compound is expressed by the molecular formula of the compound.

The molecular formula of a compound may be the empirical formula, or it may be a multiple of the empirical formula.

If the molecular formula is P4O10, this means for each for P-atoms we have 10-O atoms this is a ratio 4:10 or 1: 2.5

To find the empirical formula we divide the molecular formula by 2 what will give us P2O5

For each 2 P atoms we have 5 O-atoms. This is a ratio 1: 2.5

This is the simpliest form for the compound P4O10.

The empirical formula is P2O5 (option B)

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Answer: The volume of the gold is $75.68g/cm^3$

The mass of gold is 19.3 g

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Explanation:

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To calculate the volume of cube, we use the equation:

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Putting values in above equation, we get:

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3 years ago

Calculate ΔS° for the reaction: 4Cr(s) + 3O2(g) → 2Cr2O3(s), Substance: Cr(s) O2(g) Cr2O3(s), S°(J/K⋅mol): 23.77 205.138 81.2

Mariana [72]

<u>Answer:</u> The value of $\Delta S^o$ for the reaction is 1051.93 J/K

<u>Explanation:</u>

Entropy change is defined as the difference in entropy of all the product and the reactants each multiplied with their respective number of moles.

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

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

For the given chemical reaction:

$4Cr(s)+3O_2(g)\rightarrow 2Cr_2O_3(s)$

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

$\Delta S^o_{rxn}=[(2\times \Delta S^o_{(Cr_2O_3(s))})]-[(4\times \Delta S^o_{(Cr(s))})+(3\times \Delta S^o_{(O_2(g))})]$

We are given:

$\Delta S^o_{(Cr_2O_3(s))}=881.2J/K.mol\\\Delta S^o_{(O_2(g))}=205.13J/K.mol\\\Delta S^o_{(Cr(s))}=23.77J/K.mol$

Putting values in above equation, we get:

$\Delta S^o_{rxn}=[(2\times (881.2))]-[(4\times (23.77))+(3\times (205.13))]\\\\\Delta S^o_{rxn}=1051.93J/K$

Hence, the value of $\Delta S^o$ for the reaction is 1051.93 J/K

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