Answer:
The particles in the neutral paper can shift, causing the paper to become polarized and attracted to the rod.
Explanation:
The neutral paper has an even distribution of its electrons throughout the paper. If a charged rod is brought near the neutral paper, this can cause the electrons in the paper to shift. If the rod is negative, the electrons will be repelled from the rod and cause the molecules in the paper to have a slight positive charge on the part of the paper closest to the rod. If the rod is positive, the electrons will be attracted to the rod and cause a slight negative charge on the side of the paper closest to the rod.
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Answer:


Explanation:
first write the equilibrium equaion ,
⇄ 
assuming degree of dissociation
=1/10;
and initial concentraion of
=c;
At equlibrium ;
concentration of
![[C_3H_5O_3^{-} ]= c\alpha](https://tex.z-dn.net/?f=%5BC_3H_5O_3%5E%7B-%7D%20%20%5D%3D%20c%5Calpha)
![[H^{+}] = c\alpha](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3D%20c%5Calpha)

is very small so
can be neglected
and equation is;

= 
![P_H =- log[H^{+} ]](https://tex.z-dn.net/?f=P_H%20%3D-%20log%5BH%5E%7B%2B%7D%20%5D)





composiion ;
![c=\frac{1}{\alpha} \times [H^{+}]](https://tex.z-dn.net/?f=c%3D%5Cfrac%7B1%7D%7B%5Calpha%7D%20%5Ctimes%20%5BH%5E%7B%2B%7D%5D)
![[H^{+}] =antilog(-P_H)](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%5D%20%3Dantilog%28-P_H%29)
![[H^{+} ] =0.0014](https://tex.z-dn.net/?f=%5BH%5E%7B%2B%7D%20%5D%20%3D0.0014)


When waters salinity increases it’s freezing point will decrease. This is one of the reasons why people salt the roads after a snow storm.
Answer:
Reagents: 1)
2)
, 
Mechanism: Hydroboration
Explanation:
In this case, we have a <u>hydration of alkene</u>s reaction. But, in this example, we have an <u>anti-Markovnikov reaction</u>. In other words, the "OH" is added in the least substituted carbon. Therefore we have to choose an anti-Markovnikov reaction: <u>"hydroboration"</u>.
The <u>first step</u> of this reaction is the addition of borane (
) to the double bond. Then in the <u>second step</u>, we have the deprotonation of the hydrogen peroxide, to obtain the peroxide anion. In the <u>third step</u>, the peroxide anion attacks the molecule produced in the first step to produce a complex compound in which we have a bond "
". In <u>step number 4</u> we have the migration of the C-B bond to oxygen. Then in <u>step number 5</u>, we have the attack of
on the
to produce an alkoxide. Finally, the water molecule produce in step 2 will <u>protonate</u> the molecule to produce the alcohol.
See figure 1
I hope it helps!