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2 years ago

Bases feel slippery because they remove fats and oils from the surface of our skin.

Chemistry

2 answers:

cluponka [151]2 years ago

6 0

It is true hope it helps!

inn [45]2 years ago

6 0

Answer:

It's true

Explanation:

It's answer (a) true. Just did the quiz!

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A gas with a volume of 4.0 liters at a pressure of 205 kilopascals is allowed to expand to a volume of 12.0 liters. What will th

Ksenya-84 [330]

Boyle's law p1V1=p2V2

p2=(p1V1)/V2

p2=(205*10^3 Pa * 4*10^-3 m^3 ) / (12*10^-3 m^3)

p2= 68333 Pa

8 0

2 years ago

Which of the following would be a clue that a rock is igneous?

ryzh [129]

Answer:

Explanation:

they usually have that Crystal like thing in them... since they are formed from Lava

7 0

2 years ago

Read 2 more answers

Electrophilic addition of bromine, Br2, to alkenes yields a 1,2-dibromoalkane. The reaction proceeds through a cyclic intermedia

Taya2010 [7]

Answer:

See explanation and image attached for details

Explanation:

The reaction involves the heterolytic fission of the Br-Br bond in the bromine molecule to yield a bromine cation which attacks the but-1-ene to form a cyclic intermediate called the brominium ion. The bromine anion must now attack from the opposite face of the brominium ion due to steric clashes to form a product of a 1,2-dibromoalkane having the anti- stereochemistry.

5 0

2 years ago

The controlled Release of nuclear energy in a creator is accomplished by: allowing the critical mass of fuel to react . Controll

pishuonlain [190]

Answer:

If a chain reaction takes place, the amount of energy released would increase exponentially, so in order to control the energy release, fission must be controlled without controlling a chain reaction, so the third option is correct.

Hope this helps!

Explanation:

4 0

2 years ago

Ag2S + Al(s) = Al2S3 + Ag(s) (unbalanced)

Dovator [93]

Answer:

1. 0.97 V

2. $Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/Ag_(_s_)$

Explanation:

In this case, we can start with the half-reactions:

$Ag^+~_(_a_q_)->~Ag_(_s_)$

$Al_(_s_)~->~Al^+^3~_(_a_q_)$

With this in mind we can add the electrons:

$Ag^+~_(_a_q_)+~e^-~->~Ag_(_s_)$ Reduction

$Al_(_s_)~->~Al^+^3~_(_a_q_)+~3e^-~$ Oxidation

The reduction potential values for each half-reaction are:

$Ag_2S~+~e^-~->~Ag_(_s_)~+~S^-^2~_(_a_q_)$ - 0.69 V

$Al^+^3~_(_a_q_)+~2e^-~->~Al_(_s_)$ -1.66 V

In the aluminum half-reaction, we have an oxidation reaction, therefore we have to flip the reduction potential value:

$Al_(_s_)~->~Al^+^3~+~2e^-~$ +1.66 V

Finally, to calculate the overall potential we have to add the two values:

1.66 V - 0.69 V = 0.97 V

For the second question, we have to keep in mind that in the cell notation we put the anode (the oxidation half-reaction) in the left and the cathode (the reduction half-reaction) in the right. Additionally, we have to use "//" for the salt bridge, therefore:

$Al_(_s_)/Al^+^3~_(_a_q_)~//~Ag^+~_(_a_q_)/~Ag_(_s_)$

I hope it helps!

3 0

2 years ago