Which of the following is a homogeneous mixture?a. iced tea
b. apple pie
c. orange juice
d. vegetable soup User: A group of students want to identify a sample of matter. They observed the sample and made the observations below. • There are various components in the sample. • Different regions of the sample have different properties. Based on these observations, how will the students classify the matter?
a. heterogeneous mixture
b. homogeneous mixture
c. chemical ...

Answers

Answer 1

Answer: The answer for the first question is A. Iced Tea and C. Orange Juice. The answer for the second question is A. Heterogeneous mixture

A homogeneous mixture is a combination of 2 or more substances wherein its concentration is evenly distributed throughout the mixture. This means that you cannot identify the different components or properties of the mixture. A heterogeneous mixture, on the other hand, is a combination of 2 or more substances wherein its components are unevenly distributed. This means that its components can visibly be identified from each other.

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When should you avoid looking directly at magnesium burning?

Answers

Answer:

Magnesium also reacts with nitrogen in the air to produce some magnesium nitride (Mg3N2). Safety: Do not look directly at the burning magnesium due to the intensity of the light. A dry-powder fire extinguisher should be available. Disposal: Once cooled the solid magnesium products can be thrown in the trash.

Explanation:

Help me with 23 fast

Answers

Answer:

a salt

Explanation:

compounds are made of molecules but we need to mention the type of compound formed.

Answer:

A salt

Because compounds are made of molecules but we need to mention the type of compound formed.

77. A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25°C. The initial concentrations of Ni2+ and Zn2+ are 1.50 M and 0.100 M, respectively. a. What is the initial cell potential? b. What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M? c. What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?

Answers

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^(2+)$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^(2+)$ and $Zn^(2+)$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_([Ni^(2+)/Ni])=-0.23V$

$E^0_([Zn^(2+)/Zn])=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^(2+)+2e^-$ $E^0_([Zn^(2+)/Zn])=-0.76V$

Reaction at cathode (reduction) : $Ni^(2+)+2e^-\rightarrow Ni$ $E^0_([Ni^(2+)/Ni])=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^(2+)(aq)\rightarrow Zn^(2+)(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_(cathode)-E^o_(anode)$

$E^o=E^o_([Ni^(2+)/Ni])-E^o_([Zn^(2+)/Zn])$

$E^o=(-0.23V)-(-0.76V)=0.53V$

(a) Now we have to calculate the cell potential.

Using Nernest equation :

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)])/([Ni^(2+)])$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_(cell)$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_(cell)=0.53-(0.0592)/(2)\log ((0.100))/((1.50))$

$E_(cell)=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^(2+)$ has fallen to 0.500 M.

New concentration of $Ni^(2+)$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^(2+)$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)])/([Ni^(2+)])$

Now put all the given values in the above equation, we get:

$E_(cell)=0.53-(0.0592)/(2)\log ((1.1))/((0.500))$

$E_(cell)=0.52V$

(c) Now we have to calculate the concentrations of $Ni^(2+)$ and $Zn^(2+)$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)+x])/([Ni^(2+)-x])$

Now put all the given values in the above equation, we get:

$0.45=0.53-(0.0592)/(2)\log ((0.100+x))/((1.50-x))$

$x=1.49M$

The concentration of $Ni^(2+)$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^(2+)$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

What happens to a water bottle in a freezer and why

Answers

the water well begin to freeze and turn into soild. why? because the temperature of the the freezer is getting the water bottle cold so when that cold temperature hits that that bottle which has liquid in it so you should know that when liquid is into cold air it well so be soild so when you put the water bottle in the freezer it will be soild.
and also remember gas - liquid - soild gas change to liquid by hot air and liuid change to soild by clod air
hoped i helped

Is there any difference between carbon(iv)oxide and carbon(ii)oxide

Answers

The difference is the no. of charges present in Carbon, meaning Carbon has different charges therefore Roman numerals are used to clarify the exact number.This makes balancing equations easier.

Also their names are different.
CO4 - Carbon tetroxide
CO2 - Carbon dioxide

What is the nuclear symbol for the isotope of bromine?

Answers

The isotopes of bromine are Bromine-79 and Bromine-81. Bromine-79 has 35 protons and 44 neutrons while bromine-81 has 35 protons and 46 neutrons. They are represented as Br-79 and Br-81.

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