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Rus_ich [418]
2 years ago
8

Apple's "creative pros" and "evangelists," who help customers get the most out of their products and gain support for its produc

ts, are most likely associated with which of the following marketing mix elements?
Chemistry
1 answer:
Nesterboy [21]2 years ago
8 0

Answer:

They are most likely associated with the "Promotion" marketing mix element.

Explanation:

The marketing mix is made up of 7 elements which are:

-Product

-Price

-Promotion

-Place

-Packaging

-Positioning

-People

Promotion

Promotion essentially includes all the methods,ways and means you inform your customers about your product or services and how one can market the products and make sales from them.Apple's "creative pros" and "evangelists" basically promote Apple products by their activities since they make the customers know more about what Apple products can do thereby getting support for the Apple product which serves as a form of publicity for the products and as the publicity increases it would increase the sales of Apple products.Just as advertising serves to increase sales.

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Solid NH4HS is introduced into an evacuated flask at 24 ∘C. The following reaction takes place: NH4HS(s)⇌NH3(g)+H2S(g) At equili
Ivanshal [37]

Answer:

0.09425

Explanation:

The reactant is in solid phase and therefore has zero partial pressure.

The products have the same mole ratio (1:1) and will have the same partial pressure = 1/2 × 0.614 atm = 0.307 atm

Kp =  (NH3)(H2S) = 0.09425

4 0
3 years ago
Please help >>>>>>>>>>>
Vsevolod [243]

Answer:

Explanation:

mass of 1mole of brick =  mass of 1 brick x avagadro number

= 4 kg x 6.022 x 10^23 bricks  = 2.4 x 10^24 kg

No. of moles of bricks have a mass equal to the mass of the earth

= mass of earth / mass of 1 mole of brick

= [ 6 x 10^27 ]  /  [ 2.4 x 10^24 ]

= 2.5 x 10^3

= 2500 moles

7 0
3 years ago
The standard reduction potentials of the following half-reactions are given in Appendix E in the textbook:
german

<u>Answer:</u>

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

<u>Explanation:</u>

The substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

We are given:

Ag^++(aq.)+e^-\rightarrow Ag(s);E^o_{Ag^+/Ag}=0.799V\\\\Cu^{2+}+(aq.)+2e^-\rightarrow Cu(s);E^o_{Cu^{2+}/Cu}=0.337V\\\\Ni^{2+}(aq.)+2e^-\rightarrow Ni(s);E^o_{Ni^{2+}/Ni}=-0.28V\\\\Cr^{3+}(aq.)+3e^-\rightarrow Cr(s);E^o_{Cr^{3+}/Cr}=-0.74V

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}

  • <u>Cell having 1st and 2nd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Copper will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-0.337=0.462V

  • <u>Cell having 1st and 3rd half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-(-0.28)=1.079V

  • <u>Cell having 1st and 4th half reactions:</u>

Silver has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.799-(-0.74)=1.539V

  • <u>Cell having 2nd and 3rd half reactions:</u>

Copper has higher electrode potential. So, this will undergo reduction reaction and act as anode. Nickel will undergo oxidation reaction and act as cathode.

E^o_{cell}=0.337-(-0.28)=0.617V

  • <u>Cell having 3rd and 4th half reactions:</u>

Nickel has higher electrode potential. So, this will undergo reduction reaction and act as anode. Chromium will undergo oxidation reaction and act as cathode.

E^o_{cell}=-0.28-(-0.74)=0.46V

Hence,

<u>For 1:</u> The largest positive cell potential is of cell having 1st and 4th half reactions.

<u>For 2:</u> The standard electrode potential of the cell is 1.539 V

<u>For 3:</u> The smallest positive cell potential is of cell having 3rd and 4th half reactions. The standard electrode potential of the cell is 0.46 V

8 0
3 years ago
How can you make an unsaturated solution from a saturated solution?
riadik2000 [5.3K]

Answer:

By heating the solution a saturated solution can be changed into an unsaturated solution. Without adding any solvent it can be changed into an unsaturated solution. Example: on heating a saturated solution of sugar in high temperature, it starts dissolving.

4 0
2 years ago
Ultimately the carbon molecules in pyruvate end up as what molecules
jeyben [28]
CO2 Thats your answer
3 0
2 years ago
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