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Ksju [112]
3 years ago
8

What is the chemical reaction between acetone and styrofoam

Chemistry
1 answer:
DENIUS [597]3 years ago
6 0

Answer:

Styrofoam dissolves in acetone similar to how sugar dissolves in water.

Explanation:

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By getting ionised and losing electrons.
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Zeke is on spring break. He lives in a cold climate, so he decides to drive to a warmer climate. Before he leaves for his vacati
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The warmer temperatures at Zeke's destination caused the volume of air in his tires to increase.

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Enter the correct ground-state (or lowest energy) configuration based on the number of electrons: 1s^4 2s^4 2p^12.
Blababa [14]

Answer:

The ground state configuration is the lowest energy, most stable arrangement. An excited state configuration is a higher energy arrangement (it requires energy input to create an excited state). Valence electrons are the electrons utilised for bonding.

or the

FIGURE 5.9 The arrow shows a second way of remembering the order in which sublevels fill. Table 5.2 shows the electron configurations of the elements with atomic numbers 1 through 18.

Element Atomic number Electron configuration

sulfur 16 1s22s22p63s23p4

chlorine 17 1s22s22p63s23p5

argon 18 1s22s22p63s23p6

or the

Two electrons

Two electrons fill the 1s orbital, and the third electron then fills the 2s orbital. Its electron configuration is 1s22s1.

Explanation:

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3 years ago
A force that exists throughout the universe. Earth's will pull on you at a rate of 9.8 m/s2 if you ignore friction.
Fiesta28 [93]

Answer:

Your answer is Gravity,

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3 years ago
Use the Nernst equation to calculate the concentration of the unknown solution. Base this on your experimental voltage of 1.074
Hoochie [10]

Answer:

0.3793 M

Explanation:

The unknown metal is zinc. So the equation of the reaction is;

Zn(s) + Cu^2+(aq) -------> Zn^2+(aq) + Cu(s)

From Nernst equation;

E = E° - 0.0592/n log Q

[Cu2+] = 0.050179 M

n = 2

[Zn^2+] = ?

E = 1.074 V

E° = 0.34 - (-0.76) = 1.1 V

Substituting values;

1.074  = 1.1 - 0.0592/2 log [Zn^2+]/0.050179

1.074 - 1.1 = - 0.0592/2 log [Zn^2+]/0.050179

-0.026 = -0.0296  log [Zn^2+]/0.050179

-0.026/-0.0296 = log [Zn^2+]/0.050179

0.8784 =log [Zn^2+]/0.050179

Antilog(0.8784) = [Zn^2+]/0.050179

7.558 =  [Zn^2+]/0.050179

[Zn^2+] = 7.558 * 0.050179

[Zn^2+] = 0.3793 M

4 0
3 years ago
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