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777dan777 [17]
3 years ago
8

If a negatively charged object is brought near a neutrally charged sphere, what will happen? The electrons will move toward the

charged object. The protons will jump to the negatively charged object making both objects neutral. The electrons will jump to the charged object. The electrons will move away from the charged object.
Chemistry
2 answers:
vampirchik [111]3 years ago
7 0

The protons will jump to the negatively charged object making both objects neutral.

Sidana [21]3 years ago
5 0

Answer: Option (a) is the correct answer.

Explanation:

When a charged object is brought near a neutral atom then the neutral atom will develop an opposite charge. This process of inducing charge is known as induction.

Hence, when a negatively charged object is brought near a neutrally charged sphere then a positive charge will develop at the sphere. Thus, there will be attraction because opposite charges attract each other and like charges repel each other.

Hence, we can conclude that if a negatively charged object is brought near a neutrally charged sphere, then the electrons will move toward the charged object.

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3 years ago
If a pharmacist adds 10 ml of purified water to 30 ml of a solution having a specific gravity of 1.30, calculate the specific gr
xxMikexx [17]

The specific gravity of a sample is the ratio of the density of the sample with respect to one standard sample. The standard sample used in specific gravity calculation is water whose density is 1 g/mL. The solution having specific gravity 1.30 is the density of the sample that is 1.30 g/mL. Thus the weight of the 30 mL sample is (30×1.30) = 39 g.

Now the mass of the 10 mL of water is 10 g as density of water is 10 g/mL. Thus after addition the total mass of the solution is (39 + 10) = 49g and the volume is (30 + 10) = 40 mL. Thus the density of the mixture will be \frac{49}{40}=1.225 g/mL. Thus the specific gravity of the mixed sample will be 1.225 g/mL.

6 0
3 years ago
For the Zn - Cu^2+ voltaic cell Zn(s) + Cu^2+(aq, 1M) + Cu(s) E degree _cell = 1.10 V Given that the standard reduction potentia
Fittoniya [83]

Answer : The value of E^o_{(Cu^{2+}/Cu)} is, 0.34 V

Explanation :

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The oxidation-reduction half cell reaction will be,

Oxidation half reaction:  Zn\rightarrow Zn^{2+}+2e^-

Reduction half reaction:  Cu^{2+}+2e^-\rightarrow Cu

Oxidation reaction occurs at anode and reduction reaction occurs at cathode. That means, gold shows reduction and occurs at cathode and chromium shows oxidation and occurs at anode.

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E^o_{cell}=E^o_{(Cu^{2+}/Cu)}-E^o_{(Zn^{2+}/Zn)}

Putting values in above equation, we get:

1.10V=E^o_{(Cu^{2+}/Cu)}-(-0.76V)

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