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alexandr402 [8]
3 years ago
13

A certain skin lotion is a fine mixture of water and various oils. This lotion is cloudy and cannot be separated into oil and wa

ter by filtration. Moreover, its components do not separate when left standing. What type of mixture is it?
A.
solution
B.
suspension
C.
colloid
D.
compound
Chemistry
1 answer:
nataly862011 [7]3 years ago
4 0

Colloid

Explanation:

A lotion of this type is a typical colloid mixture.

  • A colloid is a homogeneous mixtures of two phases.
  • The dispersed phase and the dispersion medium.
  • Water is the dispersion medium and oil is the dispersed phase.

They have the following properties:

  • Liquid dispersed phase and liquid dispersion medium forms an emulsion and a lotion of this type is an example.
  • The particles are larger than those found in solutions.
  • The particles pass through ordinary filter paper.
  • They may be cloudy or clear .

learn more:

Heterogeneous mixtures brainly.com/question/1446244

#learnwithBrainly

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What volume of a 0.240 m solution of barium nitrate is needed to prepare 0.500 l a  solution that is 0.0800 m in nitrate ion?
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Formula for Barium Nitrate = Ba(NO3)2

Thus based on stoichiometry:

1 mole of Ba(NO3)2 contains 2 moles of NO3-

Therefore, concentration of nitrate ion NO3- would be = 2*0.240 = 0.480 M

Use the relation:

V1M1 = V2M2

V1 = V2M2/M1 = 0.500 L * 0.0800/0.480 = 0.0833 L

Thus, 0.0833 L or 83.3 ml solution of Ba(NO3)2 would be required.


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What properties would you expect for CaO? Select all that apply.
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Answer is:

It is likely soluble in water.

It will likely light up a bulb in a conductivity apparatus (CaO(l)).

It will likely have a high melting point.

1) Calcium oxide (CaO) has ionic bonds between calcium cations (Ca²⁺) and oxygen anions (O²⁻). It reacts with water and form calcium hydroxide:

CaO(s) + H₂O(l) → Ca(OH)₂(aq).

2) Ionic solid calcium oxide dissolved in water conduct electricty due to the dissociation Ca(OH)₂(aq) → Ca²⁺(aq) + 2OH⁻(aq).

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8 0
3 years ago
Read 2 more answers
A galvanic cell at a temperature of 25.0°C is powered by the following redox reaction:
Vlada [557]

<u>Answer:</u> The cell voltage of the given chemical reaction is 2.74 V

<u>Explanation:</u>

For the given chemical reaction:

Sn^{2+}(aq.)+Ba(s)\rightarrow Ba^{2+}(aq.)+Sn(s)

Here, gold is getting reduced because it is gaining electrons and nickel is getting oxidized because it is loosing electrons.

<u>Oxidation half reaction:</u>  Ba(s)\rightarrow Ba^{2+}(aq.)+2e^-;E^o_{Ba^{2+}/Ba}=-2.90V

<u>Reduction half reaction:</u>  Sn^{2+}(aq.)+2e^-\rightarrow Sn(s);E^o_{Sn^{2+}/Sn}=-0.14V

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the E^o_{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

E^o_{cell}=-0.14-(-2.90)=2.76V

To calculate the EMF of the cell, we use the Nernst equation, which is:

E_{cell}=E^o_{cell}-\frac{0.059}{n}\log \frac{[Sn^{2+}]}{[Ba^{2+}]}

where,

E_{cell} = electrode potential of the cell = ?V

E^o_{cell} = standard electrode potential of the cell = +2.76 V

n = number of electrons exchanged = 2

[Ba^{2+}]=5.15M

[Sn^{2+}]=1.59M

Putting values in above equation, we get:

E_{cell}=2.76-\frac{0.059}{2}\times \log(\frac{1.59}{5.15})\\\\E_{cell}=2.74V

Hence, the cell voltage of the given chemical reaction is 2.74 V

8 0
4 years ago
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