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n200080 [17]
3 years ago
5

What is the penetrating power of beta

Chemistry
1 answer:
zlopas [31]3 years ago
3 0
The penetrating power of alpha rays, beta rays, and gamma rays varies greatly. Alpha particles can be blocked by a few pieces of paper. Beta particles pass through paper but are stopped by aluminum foil. Gamma rays are the most difficult to stop and require concrete, lead, or other heavy shielding to block them.
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The correct answer is D. Meat handlers keep brain tissue from coming in contact  with other meat.

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HELPPPPP NEED FOR ESSAY Vicente is going to spend the afternoon in his yard. He has the choice to get in the pool, lay on a cott
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The higher the specific heat, the longer it will take to get hot. The coolest location for Vicente to be would be the water, because its specific heat is higher than the other locations. 

6 0
2 years ago
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If you looked at a glass containing a solution and a glass containing a suspension, how could you tell which glass contained the
marin [14]
Over time, the particles in the suspension would settle to the bottom when the movement keeping them suspended fades away. This will not happen with a solution.
6 0
2 years ago
What is the voltage of the voltaic cell Zn|Zn2+||Cu2+|Cu at 298 K if [Zn2+] = 0.2 M and [Cu2+] = 4.0 M? Cu2+ + 2e- → Cu Eo = +0.
OLga [1]

Answer : The voltage of the voltaic cell is 1.14 V

Explanation :

From the given cell representation, we conclude that

The copper will undergo reduction reaction will get reduced. Zinc will undergo oxidation reaction and will get oxidized.

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.

The overall balanced equation of the cell is,

Zn+Cu^{2+}\rightarrow Zn^{2+}+Cu

To calculate the E^o{cell} of the reaction, we use the equation:

E^o_{cell}=E^o_{cathode}-E^o_{anode}

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

Putting values in above equation, we get:

E^o_{cell}=(+0.34V)-(-0.76V)

E^o_{cell}=1.1V

Now we have to calculate the emf or voltage of the cell.

Using Nernest equation at 298 K :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Zn^{2+}]}{[Cu^{2+}]}

where,

n = number of electrons in oxidation-reduction reaction = 2

E_{cell} = ?

Now put all the given values in the above equation, we get:

E_{cell}=1.1-\frac{0.0592}{2}\log \frac{0.2}{4.0}

E_{cell}=1.14V

Therefore, the voltage of the voltaic cell is 1.14 V

5 0
2 years ago
State two reasons atom X is different than atom X+1
insens350 [35]

Explanation:

atom X

it is neutral

and may not exist independently

atom X+1

it is ion which has a charge on it positive or negative

it exists independently

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