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Inessa [10]
2 years ago
10

1. Surface area, agitation, and ____ affect the dissolving rate of a solid. 1. termperature 2. location​

Chemistry
2 answers:
Fiesta28 [93]2 years ago
4 0
I don’t understand???
mamaluj [8]2 years ago
3 0

Answer:??

Explanation:

You might be interested in
What is the standard electrode potential for a galvanic cell constructed in the appropriate way from these two half-cells?
____ [38]

E

θ

Cell

=

+

2.115

l

V

Cathode

Mg

2

+

/

Mg

Anode

Ni

2

+

/

Ni

Explanation:

Look up the reduction potential for each cell in question on a table of standard electrode potential like this one from Chemistry LibreTexts. [1]

Mg

2

+

(

a

q

)

+

2

l

e

−

→

Mg

(

s

)

−

E

θ

=

−

2.372

l

V

Ni

2

+

(

a

q

)

+

2

l

e

−

→

Ni

(

s

)

−

E

θ

=

−

0.257

l

V

The standard reduction potential

E

θ

resembles the electrode's strength as an oxidizing agent and equivalently its tendency to get reduced. The reduction potential of a Platinum-Hydrogen Electrode under standard conditions (

298

l

K

,

1.00

l

kPa

) is defined as

0

l

V

for reference. [2]

A cell with a high reduction potential indicates a strong oxidizing agent- vice versa for a cell with low reduction potentials.

Two half cells connected with an external circuit and a salt bridge make a galvanic cell; the half-cell with the higher

E

θ

and thus higher likelihood to be reduced will experience reduction and act as the cathode, whereas the half-cell with a lower

E

θ

will experience oxidation and act the anode.

E

θ

(

Ni

2

+

/

Ni

)

>

E

θ

(

Mg

2

+

/

Mg

)

Therefore in this galvanic cell, the

Ni

2

+

/

Ni

half-cell will experience reduction and act as the cathode and the

Mg

2

+

/

Mg

the anode.

The standard cell potential of a galvanic cell equals the standard reduction potential of the cathode minus that of the anode. That is:

E

θ

cell

=

E

θ

(

Cathode

)

−

E

θ

(

Anode

)

E

θ

cell

=

−

0.257

−

(

−

2.372

)

E

θ

cell

=

+

2.115

Indicating that connecting the two cells will generate a potential difference of

+

2.115

l

V

across the two cells.

5 0
2 years ago
Read 2 more answers
A hydrocarbon is a compound that contains mostly carbon and hydrogen. Calculate the percent composition (by mass) of the followi
Katena32 [7]
C7H16, where C=12.01, and H=1.01, so the weight of the molecule would be 7(12.01)+16(1.01), or 100.23. The percentage of carbon would be found by ((7*12.01)/100.23)*100=83.88% Carbon
((16*1.01)/100.23)*100=16.12% Hydrogen
7 0
3 years ago
¿Qué compone a un ecosistema?
schepotkina [342]

An ecosystem is a geographic area where plants, animals, and other organisms, as well as weather and landscape, work together to form a bubble of life. Ecosystems contain biotic or living, parts, as well as abiotic factors, or nonliving parts. Biotic factors include plants, animals, and other organism .

8 0
3 years ago
The equation below represents a double replacement/displacement reaction;
Naddik [55]
AB+XY->AY+XB
We know that the answer would be KCl because of the switching that takes place during a double displacement reaction. Just like Zn and MnO4 switched and combined, the remaining elements, K and Cl, will combine.
We know that the answer is simply KCl because both K and Cl have an ion of only +/-1, meaning when they cross, no suffixes are made, since their ions are only 1.
For example, if you combined Mg with Cl, you would get MgCl2, because Mg has an ion of +2.
I hope this helps!
4 0
3 years ago
Determine the reducing agent in the following reaction. Explain your answer. 2 Li(s) + Fe(C2H3O2)2(aq) → 2 LiC2H3O2(aq) + Fe(s)
Brilliant_brown [7]

The reducing agent in the reaction 2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s) is lithium (Li).

The general reaction is:

2Li(s) + Fe(CH₃COO)₂(aq) → 2LiCH₃COO(aq) + Fe(s)   (1)

We can write the above reaction in <u>two reactions</u>, one for oxidation and the other for reduction:

  • Oxidation reaction

Li⁰(s) → Li⁺(aq) + e⁻   (2)

  • Reduction reaction

Fe²⁺(aq) + 2e⁻ → Fe⁰(s)    (3)

We can see that Li⁰ is oxidizing to Li⁺ (by <u>losing</u> one electron) in the lithium acetate (<em>reaction 2</em>) and that Fe²⁺ in iron(II) acetate is reducing to Fe⁰ (by <u>gaining</u> two <em>electrons</em>) (<em>reaction 3</em>).  

We must remember that the reducing agent is the one that will be oxidized by <u>reducing another element</u> and that the oxidizing agent is the one that will be reduced by <u>oxidizing another species</u>.

In reaction (1), the<em> reducing agent</em> is <em>Li</em> (it is oxidizing to Li⁺), and the <em>oxidizing agent </em>is<em> Fe(CH₃COO)₂</em> (it is reducing to Fe⁰).  

Therefore, the reducing agent in reaction (1) is lithium (Li).  

 

Learn more here:

  • brainly.com/question/10547418?referrer=searchResults
  • brainly.com/question/14096111?referrer=searchResults

I hope it helps you!

3 0
2 years ago
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