CHEMISTRY COLLEGE

The activiation energy required for a chemical reaction can be decreased by? A) increasing the surface area of the reactant.
B) increasing the temperature of the reactant.
C) adding a catalyst to the reaction.
D)adding more reactant​

Answers

Answer 1
Answer:

Answer:

C)

Explanation:

C) adding a catalyst to the reaction.

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If the enthalpy change of a reaction in a flask is δh = +67 kj, what can be said about the reaction?

Answers

Answer : Normally in any chemical reaction, if the enthalpy change i.e. ΔH is positive which means it is greater than zero then it can be called as an Endothermic Reaction.

Whereas, the system under study is absorbing heat that is produced during the reaction. So if  ΔH  is found to be positive then it can be called as endothermic reaction.

The enthalpy change of the reaction indicates that it is an endothermic process.

FURTHER EXPLANATION

Enthalpy (ΔH) is the amount of heat absorbed or released in a reaction. It is based on the amount of energy needed to break the bonds and the energy released during bond formation. Enthalpy change is the difference in the enthalpy of the reactants and the products. The positive or negative sign for an enthalpy value indicates the direction of the heat flow: a positive ΔH indicates that the reaction is endothermic while a negative value for ΔH means that the reaction is exothermic.

Endothermic Reactions

Endothermic reactions are reactions that absorb heat from the surroundings to the system. This is the case when more energy is absorbed to break the bonds than is released to form the bonds. Endothermic reactions can be identified in the lab by observing if the reaction vessel becomes cooler as the reaction proceeds.

Exothermic Reactions

When the amount of energy released during bond formation is greater than the amount of energy absorbed during bond breaking, a net release of energy to the surroundings takes place and the reaction is exothermic. Exothermic reactions can be identified when the reaction vessel becomes hot as the reaction progresses.

LEARN MORE

Keywords: Endothermic, Exothermic, Enthalpy

A solution contains 0.10 M Pb2+ and 0.10 M Cu2.. Which cation will precipitate first when a solution of NazS is slowly added to the mixture? Refer to the information sheet for solubility constants. P A) Pb2+ B) Cu2+ C) impossible to tell D) both cations

Answers

Answer:

b) Cu2+

Explanation:

  • information sheet for solubility constants:

Ksp PbS = 3.4 E-28

Ksp CuS = 6.0 E-37

  • PbS ↔ Pb2+  +  S2-

∴ Ksp = 3.4 E-28 = [ Pb2+ ] * [ S2- ]

∴ [ Pb2+ ] = 0.10 M

⇒ [ S2- ] = 3.4 E-28 / 0.10 = 3.4 E-27 M

  • CuS ↔ Cu2+  +  S2-

∴ Ksp = 6.0 E-37 = [ Cu2+ ] * [ S2- ]

∴ [ Cu2+ ] = 0.10 M

⇒ [ S2- ] = 6.0 E-37 / 0.10 = 6.0 E-36 M

we have:

(1) [ S2- ] PbS >> [ S2- ] CuS

(2) Ksp PbS >> Ksp CuS

from (1) and (2) it can determined, that separation can be carried out and also the cation that precipitates first is the Cu2+

What is the rate of effusion for a gas that has a molar mass twice that of a gas thateffuses at a rate of 3.6 mol/min?

Answers

Explanation:

ans: o.o846

What volume of a solution is needed to dissolve 1.0 mol of KOH to make a solution whose pH is 13.55? A. 2.82 L
B. 12.43 L
C. 4.77 L
D. 3.35 L

Answers

Answer:

Option A. 2.82 L

Explanation:

Step 1:

Data obtained from the question.

pH = 13.55

Step 2:

Determination of the pOH of the solution.. This is illustrated below:

pH + pOH = 14

pH = 13.55

13.55 + pOH = 14

Collect like terms

pOH = 14 - 13.55

pOH = 0.45

Step 3:

Determination of the concentration of the OH ion.

This is illustrated below:

pOH = - Log [OH-]

pOH = 0.45

0.45 = - Log [OH-]

- 0.45 = Log [OH-]

[OH-] = antilog (- 0.45)

[OH-] = 0.355 M

Step 4:

Determination of the molarity of KOH. This is illustrated below:

First, we'll write the dissociation equation of KOH as follow:

KOH —> K+(aq) + OH-(aq)

From the balanced equation above,

1 mole of KOH produced 1 mole of OH-.

Therefore, 0.355 M of KOH will definitely produce 0.355 M of OH-.

The molarity of KOH is 0.355 M

Step 5:

Determination of the volume of the solution needed to dissolve 1 mole of KOH. This is illustrated below:

Mole of KOH = 1 mole

Molarity of KOH = 0.355 M

Volume =?

Molarity = mole /Volume

Volume = mole /Molarity

Volume = 1/0.355 M

Volume = 2.82L

Is 3.5:Aqueous solutions of iron(III) bromide and ammonium carbonate react to form a precipitate. Answer the follwing
questions with regards to this reaction.
a) Write the molecular equation for this reaction by
Translating the two reactants into their chemical formulae.
Predict the products.
Label all the states.
Balance the reaction.

Answers

Answer:

2 FeBr₃(aq) + 3 (NH₄)₂CO₃(aq) = Fe₂(CO₃)₃(s) + 6 NH₄Br(aq)

Explanation:

Aqueous solutions of iron(III) bromide and ammonium carbonate react. This is a double displacement reaction that gives place to ammonium bromide and iron (III) carbonate. Iron (III) carbonate is insoluble so it precipitates. The corresponding molecular equation is:

2 FeBr₃(aq) + 3 (NH₄)₂CO₃(aq) = Fe₂(CO₃)₃(s) + 6 NH₄Br(aq)

A binary feed mixture contains 40 mol% hexane (A) and 60 mol% toluene (B) is to be separated continuously into two products D (distillate) and B (bottoms) in a distillation unit. Distillate D is 90 mol% hexane and the bottoms B is 90 mol% toluene. Using a feed flow rate of 100 lbmoh as basis, compute the flow rates of products B and D in: (a) lbmol/h, and (b) kmol/h.

Answers

Answer:

a) D = 33.44 Lbmol/h

⇒ B = 62.56 Lbmol/h

b) D = 16.848 Kmol/h

⇒ B = 28.152 Kmol/h

Explanation:

global balance:

  • F = D + B........................(1)

∴ F = 100 Lbmol/h

balance per component:

A: 0.4*F = 0.9*D + 0.1*B = 0.4*100 = 40 Lbmol/h..............(2)

B: 0.6*F = 0.1*D + 0.9*B = 0.6*100 = 60 Lbmol/h..............(3)

from (2):

⇒ 0.9*D = 40 - 0.1*B

⇒ D = ( 40 - 0.1*B ) / 0.9............(4)

(4) in (3):

⇒ 0.1*((40-0.1*B)/0.9) + 0.9*B = 60

⇒ B = 62.56 Lbmol/h............(5)

(5) in (1):

⇒ D = 100 - B

⇒ D = 37.44 Lbmol/h

∴ Lbmol = 0.45 Kmol

⇒ B = 62.56 Lbmol/h * ( 0.45 Kmol/ Lbmol ) = 28.152 Kmol/h

⇒ D = 37.44 Lbmol/h * ( 0.45 Kmol/h ) = 16.848 Kmol/h
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