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2 years ago

A mixture of 117.9 g of P and 121.8 g of O, reacts completely to form P4O6 and P4O10. Find the masses of P4O6 and P4O10

Chemistry

1 answer:

Alex17521 [72]2 years ago

4 0

Answer:

Mass of P4O6=103.4

P4O10=133.48

Explanation:

Balanced reaction is:

8P +8 $O_{2}$ ⇒ $P_{4} O_{6}$ + $P_{4} O_{10}$

Both reactant completely vanishes as equivalent of bot are equal.

Moles of P= $\frac{117.9}{31}$ =3.80

Moles of $O_{2}$ = $\frac{117.9}{31}$ =3.80

No. of moles of formed product are equal and is $\frac{1}{8}$ th of mole of any of reactant.

Thus weight of $P_{4} O_{6}$ = $\frac{3.80}{8}$ ×220 ≈103.41

weight of $P_{4} O_{10}$ = $\frac{3.80}{8}$ ×284 ≈133.48

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Answer:

The density of the ideal gas is directly proportional to its molar mass.

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Density is a scalar quantity that is denoted by the symbol ρ (rho). It is defined as the ratio of the mass (m) of the given sample and the total volume (V) of the sample.

$\rho = \frac{m}{V}$ ......equation (1)

According to the ideal gas law for ideal gas:

$PV = nRT$ ......equation (2)

Here, V is the volume of gas, P is the pressure of gas, T is the absolute temperature, R is Gas constant and n is the number of moles of gas

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$\frac{PM}{RT}= \frac{m}{V} = \rho$

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3 years ago

Match the following aqueous solutions with the appropriate letter from the column on the right.1. 0.19 m AgNO3 2. 0.17 m CrSO4 3

vichka [17]

Answer:

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0.19 m of $AgNO_3$ → Second Highest boiling point

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Explanation:

Elevation in boiling is given by :

$\Delta T_b=i\times k_b\times m$

Where :

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Elevation is boiling point of solution:

$\Delta T_b=2\times k_b\times 0.19 m$

$\Delta T_b=0.38 m\times k_b$

2) 0.17 m of $CrSO_4$

$CrSO_4\rightarrow Cr^{2+}+SO_4^{2-}$

i = 2 (electrolyte)

Molality of the solution = 0.17

Elevation is boiling point solution :

$\Delta T_b=2\times k_b\times 0.17 m$

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