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zlopas [31]
4 years ago
13

How does the mass of a planet or object effect its gravitational pull?

Chemistry
1 answer:
mrs_skeptik [129]4 years ago
6 0
The gravitational "range" will be the same whether or not there is mass in the way if gravity will have an effect on an object for 100 miles but there's 50 miles of land in the way, it will still affect the 100 mile radius. so it really does if the range is 4000 miles and the planet is 2000 miles thick then the pull will be weaker than that of a planet 1000 miles thick bc you are closer to the source of gravity
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The following substances dissolve when added to water. Classify the substances according to the strongest solute-solvent interac
Norma-Jean [14]

Answer:

CuCl2-Ion-dipole forces

CuSO4-Ion-dipole forces

NH3-Dipole-dipole forces

CH3OH-Dipole-dipole forces

Explanation:

Water consists of a dipole. The water molecule contains a positive end and a negative end. The positive ion attracts the negative dipole of water while the positive dipole in water interacts with the negative ion of an ionic substance. This explains the dissolution of ionic substances in water.

Copper II chloride and copper sulphate are ionic substances hence they dissolve by the mechanism described above.

Molecules consisting of dipoles dissolves by interaction of the molecule's dipoles with the dipoles in water. For example, methanol interacts with water through hydrogen bonding which is involves molecular dipoles

3 0
3 years ago
Select all that apply.When the products of a reaction are hotter than the reactants
andrezito [222]
When the products of a reaction are hotter than the reactants, an exothermic reaction is happening. An exothermic reaction is a reaction that releases energy to the surroundings. The energy released should be more than what is absorbed in order to maintain the reaction.
8 0
3 years ago
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

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