How do simple distillation and fractional distillation differ?

1 answer:

5 0

Fractional distillation is essentially the same as simple distillationexcept that a fractionating column is placed between the boiling flask and the condenser. The fractionating column is usually filled with glass or plastic beads. These beads improve the separation between the liquids being distilled.

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When will Le Chatelier's principle come into effect?

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If u disturbed equilibrium position then this principal comes into effect deciding how to counteract the disturbance.

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3 years ago

The amount of force needed to keep a 10 kg box moving at a constant

boyakko [2]

Answer:

Explanation:

Frictional force is a force that opposes the motion of body. It is because of friction that motion can be controlled and even walking is made possible.

Now, on a frictionless surface, there is no opposition to motion. Therefore, to keep the body moving there is no force required.

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3 years ago

How many are molecules ( or formula) in each sample?

andre [41]

Answer:

$4.010 \times 10^{25} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$
$16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

Explanation:

Number of molecules for $55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

$\text { Firstly molar mass is calculated of } \mathrm{NaHCO}_{3}:$

Atomic mass of Na + H + C + 3(O) = 22.99 + 1.008 + 12.01 + 3 × 16.00 = 84.00 g/mol

$\text { Number of molecules of } \mathrm{NaHCO}_{3} \text { in } 55.93 \text { kg are as follows: }$

$55.93 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} \mathrm{NaHCO}_{3}}{84.00 \mathrm{gm} \mathrm{NaHCO}_{3}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number }\right)$

$=4.010 \times 10^{26} \text { molecules of } \mathrm{NaHCO}_{3} \text { present in } 55.93 \mathrm{kg} \text { of } \mathrm{NaHCO}_{3}$

Number of molecules for for $\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

$\text { Firstly molar mass is calculated of } \mathrm{Na}_{3} \mathrm{PO}_{4}$

= Atomic mass of 3(Na) + P + 4(O)

= 3(22.99) + 30.97 + 4(16.00) = 163.94 g/mol

$459 \times\left(10^{3} \mathrm{gm}\right) \times \frac{1 \mathrm{mol} N a_{3} P O_{4}}{163.94 \mathrm{gm} N a_{3} P O_{4}} \times\left(6.022 \times 10^{23} \mathrm{molecules} \text { i.e Avogadro number) } / 1 \mathrm{mol}\right.$

$=16.86 \times 10^{26} \text { molecules of } \mathrm{Na}_{3} \mathrm{PO}_{4} \text { present in } 459 \mathrm{kg}\left(4.59 \times 10^{5} \mathrm{gm}\right) \text { of } \mathrm{Na}_{3} \mathrm{PO}_{4}$