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3 years ago

A compound is 44.7% P and the rest O. Determine its emperical formula.

Chemistry

1 answer:

svet-max [94.6K]3 years ago

6 0

Answer:

P₅O₁₂

Explanation:

Assume that you have 100 g of the compound.

Then you have 44.7 g P and 55.3 g O.

1. Calculate the moles of each atom

Moles of P = 44.7 × 1/30.97 = 1.443 mol Al

Moles of O = 55.3 × 1/16.00 = 3.456 mol O

2. Calculate the molar ratios.

P: 1.443/1.443 = 1

O: 3.456/1.443 = 2.395

3. Multiply by a number to make the ratio close to an integer

P: 5 × 1 = 5

O: 5 × 2.395 = 11.97

3. Determine the empirical formula

Round off all numbers to the closest integer.

P: 5

O: 12

The empirical formula is P₅O₁₂.

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Therefore, when a 100 g of given pure metal in solid state is heated at its exact melting point which is $215^{o}C$ then some of the solid will change into liquid state but the temperature will remains the same.

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3 years ago

Using any data you can find in the ALEKS Data resource, calculate the equilibrium constant K at 30.0 °C for the following reacti

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Answer : The value of $K$ for this reaction is, $2.6\times 10^{15}$

Explanation :

The given chemical reaction is:

$CH_3OH(g)+CO(g)\rightarrow HCH_3CO_2(g)$

Now we have to calculate value of $(\Delta G^o)$ .

$\Delta G^o=G_f_{product}-G_f_{reactant}$

$\Delta G^o=[n_{HCH_3CO_2(g)}\times \Delta G^0_{(HCH_3CO_2(g))}]-[n_{CH_3OH(g)}\times \Delta G^0_{(CH_3OH(g))}+n_{CO(g)}\times \Delta G^0_{(CO(g))}]$

where,

$\Delta G^o$ = Gibbs free energy of reaction = ?

n = number of moles

$\Delta G^0_{(HCH_3CO_2(g))}$ = -389.8 kJ/mol

$\Delta G^0_{(CH_3OH(g))}$ = -161.96 kJ/mol

$\Delta G^0_{(CO(g))}$ = -137.2 kJ/mol

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

$\Delta G^o=[1mole\times (-389.8kJ/mol)]-[1mole\times (-163.2kJ/mol)+1mole\times (-137.2kJ/mol)]$

$\Delta G^o=-89.4kJ/mol$

The relation between the equilibrium constant and standard Gibbs, free energy is:

$\Delta G^o=-RT\times \ln K$

where,

$\Delta G^o$ = standard Gibbs, free energy = -89.4 kJ/mol = -89400 J/mol

R = gas constant = 8.314 J/L.atm

T = temperature = $30.0^oC=273+30.0=303K$

$K$ = equilibrium constant = ?

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

$-89400J/mol=-(8.314J/L.atm)\times (303K)\times \ln K$

$K=2.6\times 10^{15}$

Thus, the value of $K$ for this reaction is, $2.6\times 10^{15}$

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4 years ago

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3 years ago

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Answer:

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From hook's law,

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Make k the subject of the equation,

k = F/e ............................ Equation 2

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