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qwelly [4]
3 years ago
6

Why are we pulled to the center of the Earth and not into the much larger sun?

Chemistry
2 answers:
katrin [286]3 years ago
8 0

Answer:

The sun's gravity pulls on the earth, and the earth pulls back on the sun at the same time. This is why the center of the solar system is not the center of the sun. As one gravitational body gets bigger than the other, it circles closer to the center of the system (shown in red).

Explanation:

hope this helps...

solong [7]3 years ago
7 0

Answer:

the atmosphere

Explanation:

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Answer:

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Explanation:

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Which best describes the field of chemistry?
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Fundamentally, chemistry is the study of matter and change. The way that chemists study matter and change and the types of systems that are studied varies dramatically. Traditionally, chemistry has been broken into five main subdisciplines: Organic, Analytical, Physical, Inorganic, and Biochemistry.

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The specific heat of gold is 0.031 calories/gram°C and the specific heat of silver is 0.057 calories/gram°C. If equal amounts of
Vikki [24]
I think the substance that will heat up faster would be the silver metal since it has a higher heat capacity. Heat capacity is the amount of heat needed to raise the temperature of the system into one degree. Heat capacity and heat energy is directly related so higher value of heat capacity would lead to higher heat energy.
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3 years ago
Read 2 more answers
Consider the reaction of metallic copper with iron(!!) to give copper(ll) and ironin 0.77V Fe* (aq) + e-Fe (aq) Cup (aq) + 2e --
frosja888 [35]

Answer :

(a) The anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) The emf of cell potential is 1.022 V

Explanation :

(a) The standard reduction potentials for iron and copper are:

E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Cu^{2+}/Cu)}=0.34V

In the voltaic cell, the oxidation occurs at an anode which is a negative electrode and the reduction occurs at the cathode which is a positive electrode.

From the standard reduction potentials we conclude that, the substance having highest positive E^o potential will always get reduced and will undergo reduction reaction.

So, iron will undergo reduction reaction will get reduced. Copper will undergo oxidation reaction and will get oxidized.

The given cell reactions are:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

Thus, the anode and cathode will be E^o_{(Cu^{2+}/Cu)}

and E^o_{(Fe^{3+}/Fe^{2+})} respectively.

(b) Now we have to calculate the potential of a cell.

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  Fe^{3+}+1e^-\rightarrow Fe^{2+}

In order to balance that electrons, we will multiple the reduction reaction by 2, we get:

Oxidation half reaction (anode):  Cu\rightarrow Cu^{2+}+2e^-

Reduction half reaction (cathode):  2Fe^{3+}+2e^-\rightarrow 2Fe^{2+}

The overall cell reaction will be,

2Fe^{3+}+Cu\rightarrow Cu^{2+}+2Fe^{2+}

E^o_{[Fe^{3+}/Fe^{2+}]}=2\times 0.77V=1.54V

E^o_{[Cu^{2+}/Cu]}=0.34V

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

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

E^o=E^o_{[Fe^{3+}/Fe^{2+}]}-E^o_{[Cu^{2+}/Cu]}

E^o=1.54V-(0.34V)=1.20V

Now we have to calculate the cell potential.

Using Nernest equation :

E_{cell}=E^o_{cell}-\frac{0.0592}{n}\log \frac{[Fe^{2+}]^2[Cu^{2+}]}{[Fe^{3+}]^2}

where,

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

E_{cell} = emf of the cell = ?

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

E_{cell}=1.20-\frac{0.0592}{2}\log \frac{(0.20)^2(0.25)}{(0.0001)^2}

E_{cell}=1.022V

Therefore, the emf of cell potential is 1.022 V

5 0
3 years ago
Please help I have no idea tbh
coldgirl [10]

Answer: I believe the answer is B. Sorry if i'm wrong

Explanation:

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