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Rama09 [41]
3 years ago
11

The corner where a perfume is initally sprayed has:

Chemistry
2 answers:
BartSMP [9]3 years ago
8 0
Answer D; if you spray perfume into a corner, it will eventually diffuse, but initially it has a higher concentration than all other corners of that room.
Fudgin [204]3 years ago
4 0
D. Since its initially sprayed there, it takes awhile to move to the areas of lower concentration, so for a while, it will have the highest
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A galvanic (voltaic) cell has the generic metals X and Y as electrodes. X is more reactive than Y , that is, X more readily reac
Margaret [11]

Answer:

(1) cathode: Y

(2) anode X

(3) electrons in the wire flow toward: Y

(4) electrons in the wire flow away from: X

(5) anions from the salt bridge flow toward X

(6) cations from the salt bridge flow toward Y

(7) gains mass: Y

(8) looses mass X

Explanation:

The voltaic cell uses two different metal electrodes, each in an electrolyte solution. The anode will undergo oxidation and the cathode will undergo reduction. The metal of the anode will oxidize, going from an oxidation state of 0 (in the solid form) to a positive oxidation state, and it will become an ion. At the cathode, the metal ion in the solution will accept one or more electrons from the cathode, and the ion’s oxidation state will reduce to 0. This forms a solid metal that deposits on the cathode. The two electrodes must be electrically connected to each other, allowing for a flow of electrons that leave the metal of the anode and flow through this connection to the ions at the surface of the cathode. This flow of electrons is an electrical current that can be used to do work, such as turn a motor or power a light.

7 0
3 years ago
What is one way to test whether an unknown solution is acidic or basic?
sertanlavr [38]

Answer: There are several ways. The first that comes to mind is a pH meter. A pH electrode Is lowered into the solution, and (Assuming) the pH Meter has been properly calibrated, and the temperature of the solution is set to the calibration of the Meter, the pH can be read directly from an analogue scale or digital readout. Below 7 is acidic, 7 is Neutral, (like Pure Water), and over 7 is Alkaline, or Basic.

A useful, but less accurate method is the use of any number of “pH Indicator Solutions”, which are essentially a type of various colored dyes that change color within differing pH ranges. Usually, if the pH is unknown, a small amount of solution is removed from the container and tested separately - in a “well plate”, or similar method.

These types of dyes, or Indicator Solutions, can be dried upon strips of “pH indicator Paper”, which, depending upon the type can be very useful when carrying out more precisely arrived at pH tests like Titration.

Just to see if a solution is “Acid” or “Base”, Litmus paper is used; “a Red color shows Acidity, and a Blue color, a Base”; ergo, “An Acid Solution will turn Litmus Paper, Red”.

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3 years ago
Please help me with letter a question
Colt1911 [192]
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7 0
3 years ago
Compute the percent ionic character of the interatomic bonds for the following compounds : a. TiO2 b. ZnTe c. CsCld. InSb e. MgC
sesenic [268]

Answer:

a. 63.2%

b. 11.7%

c. 73.3%

d. 0.995%

e. 55.5%

Explanation:

An ionic compound is a compound that is formed by ions, so one of the elements must donate electrons (which is the cation, the positive ion), and the other will receive these electrons (which is the anion, the negative ion).

The power of an element has to attract the electrons is called electronegativity, and so, as higher is the difference of electronegative of the elements, it is more probable that one of them will "still" the electrons and will form an ionic compound. The percent of this ionic character can be found by the Pauling's equation:

%IC = (1 - e^{-0.25*(x_A - x_B)^2}) *100%

Where x_A - x_B is the electronegativity difference of the elements. Thus, consulting an electronegativity table:

a. x_{Ti} = 1.5

x_{O} = 3.5

%IC = (1 - e^{-0.25*(3.5 - 1.5)^2})*100%

%IC = 63.2%

b. x_{Zn} = 1.6

x_{Te} = 2.1

%IC = (1 - e^{-0.25*(2.1 - 1.6)^2})*100%

%IC = 11.7%

c. x_{Cs} = 0.7

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 0.7)^2})*100%

%IC = 73.3%

d. x_{In} = 1.7

x_{Sb} = 1.9

%IC = (1 - e^{-0.25*(1.9 - 1.7)^2})*100%

%IC = 0.995 %

e. x_{Mg} = 1.2

x_{Cl} = 3.0

%IC = (1 - e^{-0.25*(3.0 - 1.2)^2})*100%

%IC = 55.5%

4 0
3 years ago
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