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pickupchik [31]
2 years ago
15

What is the similarity between pure substances and mixtures?

Chemistry
2 answers:
Firlakuza [10]2 years ago
6 0
Mixtures are made up of 2 or more pure substances. pure substance is just its self and a mixture is multiple substances.
yuradex [85]2 years ago
4 0

Answer:

Mixtures and pure substances are alike because mixtures are made up of two or more pure substances. This means that where pure substances have a single set of properties, mixtures may have two or more sets of the same properties, based on the pure substances that make up the mixture.

Explanation:

Hope it helps, Good luck And mark as BRAINLIEST

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The equilibrium constant has been estimated to be 0.12 at 25 °C. If you had originally placed 0.069 mol of cyclohexane in a 2.8
scZoUnD [109]

Answer: Concentrations of cyclohexane and methylcyclopentane at equilibrium are 0.0223 M and 0.0027 M respectively

Explanation:

Moles of cyclohexane = 0.069 mole

Volume of solution = 2.8 L

Initial concentration of cyclohexane =\frac{moles}{Volume}=\frac{0.069}{2.8}=0.025M

The given balanced equilibrium reaction is,

                            cyclohexane  ⇔  methylcyclopentane

Initial conc.                 0.025 M           0

At eqm. conc.       (0.025-x)M       (x) M

The expression for equilibrium constant for this reaction will be,

K= methylcyclopentane / cyclohexane

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

0.12=\frac{(x)}{(0.025-x)}

By solving the term 'x', we get :

x =  0.0027

Concentration of cyclohexane at equilibrium = (0.025-x ) M = (0.025-0.0027) M = 0.0223 M

Concentration of methylcyclopentane at equilibrium = (x ) M = (0.0027) M

4 0
3 years ago
The chemical formula of lead sulfide tells you that it contains<br>​
ra1l [238]

Answer:

sulfur oxygen

Explanation:

because its contain ide

7 0
3 years ago
What is an open-loop system?
Naddika [18.5K]

A.  A group of related objects that do not send out or receive feedback and cannot modify themselves

Explanation:

An open-loop system is a a group of related objects or systems that cannot send out or receive feedback and modify themselves.

  • It is a non-feedback system.
  • In this system, the output control system has no effect on whatever input that is fed into the system.
  • Output and input in such systems are independent of one another.
  • The input and output has no control whatever on each other.

learn more:

Computer programs brainly.com/question/9409412

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5 0
2 years ago
What is the oxidizing agent in the following equation? Al (s) + 3 Ag+ (aq) produces Al^+3 (aq) + 3 Ag (s)
Keith_Richards [23]

Answer:

Ag is the oxidizing agent

Explanation:

oxidizing agent in the following equation?

Al (s) + 3 Ag+ (aq) = Al+3 (aq) + 3 Ag (s)

Left side

Al = 1

Ag = 3

Right Side

Al = 1

Ag = 3

So it's balanced already good.

Define

oxidizing agent = An oxidizing agent is the substance that gains electrons and is reduced in a chemical reaction.

Al is the reducing agent.

Ag is the oxidizing agent

7 0
3 years ago
Calculate the standard reaction Gibbs free energy for the following cell reactions: (a) 2 Ce41(aq) 1 3 I2(aq) S 2 Ce31(aq) 1 I32
Law Incorporation [45]

<u>Answer:</u>

<u>For a:</u> The standard Gibbs free energy of the reaction is -347.4 kJ

<u>For b:</u> The standard Gibbs free energy of the reaction is 746.91 kJ

<u>Explanation:</u>

Relationship between standard Gibbs free energy and standard electrode potential follows:

\Delta G^o=-nFE^o_{cell}           ............(1)

  • <u>For a:</u>

The given chemical equation follows:

2Ce^{4+}(aq.)+3I^{-}(aq.)\rightarrow 2Ce^{3+}(aq.)+I_3^-(aq.)

<u>Oxidation half reaction:</u>   Ce^{4+}(aq.)\rightarrow Ce^{3+}(aq.)+e^-       ( × 2)

<u>Reduction half reaction:</u>   3I^_(aq.)+2e^-\rightarrow I_3^-(aq.)

We are given:

n=2\\E^o_{cell}=+1.08V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-2\times 96500\times (+1.80)=-347,400J=-347.4kJ

Hence, the standard Gibbs free energy of the reaction is -347.4 kJ

  • <u>For b:</u>

The given chemical equation follows:

6Fe^{3+}(aq.)+2Cr^{3+}+7H_2O(l)(aq.)\rightarrow 6Fe^{2+}(aq.)+Cr_2O_7^{2-}(aq.)+14H^+(aq.)

<u>Oxidation half reaction:</u>   Fe^{3+}(aq.)\rightarrow Fe^{2+}(aq.)+e^-       ( × 6)

<u>Reduction half reaction:</u>   2Cr^{2+}(aq.)+7H_2O(l)+6e^-\rightarrow Cr_2O_7^{2-}(aq.)+14H^+(aq.)

We are given:

n=6\\E^o_{cell}=-1.29V\\F=96500

Putting values in equation 1, we get:

\Delta G^o=-6\times 96500\times (-1.29)=746,910J=746.91kJ

Hence, the standard Gibbs free energy of the reaction is 746.91 kJ

7 0
3 years ago
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