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1 year ago

In a simple distillation setup, what is the sequence of equipment from the bench top to the round bottom flask

Chemistry

1 answer:

Julli [10]1 year ago

3 0

In a simple distillation setup, the sequence of equipment from the bench top to the round bottom flask is:

Thermometer
Distillation flask

Liebig condenser

Round bottom flask
Bunsen burner

<h3>What is Distillation?</h3>

This is the process in which a mixture is separated through selective boiling and condensation.

The distillation flask and liebig condenser are usually located above the round bottom flask in the set up.

Read more about Distillation here brainly.com/question/24553469

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You might be interested in

Which of these statements is not true about chemical reaction rates

grin007 [14]

Answer: The statement (B) is not true about chemical reactions.

Explanation:

A chemical reaction rate is affected by the several factors few of which are temperature, concentration of reactants, surface area etc.

In a chemical reaction, if temperature is increased then the rate of reaction will increase because it will increase the average kinetic energy of the reactant molecules. Thus, large number of molecules will have minimum energy required for an effective collision.

It is known that increasing the amount of reactants will increase the rate of reaction.

Therefore, rate of reaction will change if concentration or temperature is changed.

Hence, the statement (B) is not true about chemical reactions.

6 0

3 years ago

Read 2 more answers

When 125 mL of 0.150 M Pb(NO3)2 is mixed with 145 mL of 0.200 M KBr, 4.92 g of PbBr2 is collected. Calculate the percent yield.

Semenov [28]

Answer:

Y = 92.5 %

Explanation:

Hello there!

In this case, since the reaction between lead (II) nitrate and potassium bromide is:

$Pb(NO_3)_2+2KBr\rightarrow PbBr_2+2KNO_3$

Exhibits a 1:2 mole ratio of the former to the later, we can calculate the moles of lead (II) bromide product to figure out the limiting reactant:

$0.125L*0.150\frac{molPb(NO_3)_2}{L} *\frac{1molPbBr_2}{1molPb(NO_3)_2} =0.01875molPbBr_2\\\\0.145L*0.200\frac{molKBr}{L} *\frac{1molPbBr_2}{2molKBr} =0.0145molPbBr_2$

Thus, the limiting reactant is the KBr as it yields the fewest moles of PbBr2 product. Afterwards, we calculate the mass of product by using its molar mass:

$0.0145molPbBr_2*\frac{367.01gPbBr_2}{1molPbBr_2} =5.32gPbBr_2$

And the resulting percent yield:

$Y=\frac{4.92g}{5.32g} *100\%\\\\Y=92.5\%$

Regards!

4 0

3 years ago

djyliett [7]

Answer:

Explanation:

1. The amount of CaCO3 must be so small that

is less than KP when the CaCO3 has completely decomposed. In other words, the starting amount of CaCO3 cannot completely generate the full

required for equilibrium.

3. The change in enthalpy may be used. If the reaction is exothermic, the heat produced can be thought of as a product. If the reaction is endothermic the heat added can be thought of as a reactant. Additional heat would shift an exothermic reaction back to the reactants but would shift an endothermic reaction to the products. Cooling an exothermic reaction causes the reaction to shift toward the product side; cooling an endothermic reaction would cause it to shift to the reactants’ side.

5. No, it is not at equilibrium. Because the system is not confined, products continuously escape from the region of the flame; reactants are also added continuously from the burner and surrounding atmosphere.

7. Add N2; add H2; decrease the container volume; heat the mixture.

9. (a) ΔT increase = shift right, ΔP increase = shift left; (b) ΔT increase = shift right, ΔP increase = no effect; (c) ΔT increase = shift left, ΔP increase = shift left; (d) ΔT increase = shift left, ΔP increase = shift right.

11. (a)

[

]

[

]

[

]

; (b) [H2] increases, [CO] decreases, [CH3OH] increases; (c), [H2] increases, [CO] decreases, [CH3OH] decreases; (d), [H2] increases, [CO] increases, [CH3OH] increases; (e), [H2] increases, [CO] increases, [CH3OH] decreases; (f), no changes.

13. (a)

[

]

[

]

[

]

; (b) [H2O] no change, [CO] no change, [H2] no change; (c) [H2O] decreases, [CO] decreases, [H2] decreases; (d) [H2O] increases, [CO] increases, [H2] decreases; (f) [H2O] decreases, [CO] increases, [H2] increases. In (b), (c), (d), and (e), the mass of carbon will change, but its concentration (activity) will not change.

15. Only (b)

17. Add NaCl or some other salt that produces Cl− to the solution. Cooling the solution forces the equilibrium to the right, precipitating more AgCl(s).

19. (a)

Hope this helps :)

3 0

3 years ago

Between Lab Period 1 and Lab Period 2, design a separation scheme for all 4 cations. Use the results of your preliminary tests a

fiasKO [112]

Answer:

SEPARATION SCHEME FOR CATIONS

GIVEN CATIONS : $Ag^{+} \ , Fe^{3+} , Cu^{2+}, Ni^{2+}$

Step 1: Add $6mol/dm^3$ of $HCl$ to the mixture solution

Result : This would cause a precipitate of $AgCl$ to be formed

Reaction : $Ag^{+} _{(aq)} + Cl^{-} _{(aq)} ---------> AgCl(ppt)$

Step 2 : Next is to remove the precipitate and add $H_2S$ to the remaining

solution in the presence of $0.2 \ mol/dm^3$ of HCl

Result : This would cause a precipitate of $CuS$ to be formed

Reaction : $Cu^{2+}_{(aq)} + S^{2-}_{(aq)} ------> Cu_2S(ppt)$

Step 3: Next remove the precipitate then add $6 \ mol/dm^3$ of aqueous

$NH_3 (NH_3 \cdot H_2 O)$ , process the solution in a centrifuge,when the

process is done then sort out the precipitate from the solution

Now this precipitate is $Fe(OH)_3$ and the remaining solution

contains $(Ni (NH_3)_6)$

Next take out the precipitate to a different beaker and add HCl

to it this will dissolve it, then add a drop of $NH_4SCN$ this will

form a precipitate $Fe(SCN)_{6}^{3-}$ which will have the color of

blood indicating the presence of $Fe^{3+}$

Reaction : $F^{3+}_{(aq)} + 30H^-_{(aq)} --------->Fe(OH)_3_{(aq)}$

$Fe (OH)_{(s)} _3 + 3H^{+}_{aq} -------> Fe^{3+}_{aq} + 3H_2O_{(l)}$

$Fe^{3+} + 6SCN^{-} -----> Fe(SCN)_6 ^{3-}$

Now the remaining mixture contains $Ni^{2+}$

Explanation:

6 0

3 years ago

Rate law equation The rate of a chemical reaction depends on the concentrations of the reactants. For the general reaction betwe

jekas [21]

<u>Answer:</u> The reaction order with respect to A is 'm'

<u>Explanation:</u>

Order of the reaction is the sum of the concentration of terms on which the rate of the reaction actually depends. It is equal to the sum of the exponents of the molar concentration in the rate law expression.

Elementary reactions the reactions for which the order of the reaction is same as its molecularity and order with respect to each reactant is equal to its stoichiometric coefficient as represented in the balanced chemical equation.

The given chemical equation follows:

$aA+bB\rightleftharpoons cC+dD$

The rate of the above reaction is given to us as:

$Rate=k[A]^m[B]^n$

In the above rate law expression, the order with respect to the reactants is not equal to the stoichiometric coefficients. Thus, it is not an elementary reaction.

Order with respect to reactant A = m

Order with respect to reactant B = n

Hence, the reaction order with respect to A is 'm'

8 0

3 years ago